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COMMON  SCHOOL  ALGEBRA. 


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^THOMAS    SHERWIN,    A.M., 

rRINCIPAL  OF  THE  ENQLISH  HieH  SCHOOL,  BOSTON;    AUTHOR  Of 
-^  'ELEMENIART  TREATISE  ON  ALaEBRA.' 

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BOSTON: 

TAG  GAUD    &    TFIOMPSON, 

29     CORNHILL, 
1867. 


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Entered  according  to  Act  of  Congress,  in  the  year  1845,  by 

THOMAS    SHERWIN, 

kn  the  Clerk's  Office  of  the  District  Courtof  the  District  of  Massachusetts 


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STEREOTYPED  AT  THE 
BCMITON  TYPE  AND  STEREOTYPE  FOUNDRT. 


PREFACE- 


The  great  difficulty,  in  the  study  of  Algebra,  is  to 
attain  a  clear  comprehension  of  the  earliest  steps. 
The  first  principles  should,  therefore,  be  commu- 
nicated to  the  learner  gradually,  and  in  the  most 
simple  and  intelligible  manner. 

Experience  proves  that  these  principles  are  most 
successfully  taught  by  means  of  easy  problems. 
But  even  when  this  mode  is  pursued,  a  majority  of 
pupils  find  trouble  in  expressing  algebraically  the 
conditions  of  the  problems.  The  author  has,  there- 
fore,' placed  at  the  commencement  of  his  work  a 
series  of  introductory  exercises,  designed  to  famil- 
iarize the  learner  with  representing  quantities  and 
performing  the  simplest  algebraic  processes,  also  to 
prepare  him  for  putting  problems  into  equations. 

These  introductory  exercises,  which  were  written 
about  three  years  since,  were  shown  to  several 
excellent  teachers,  and  received  their  approbation. 
They  were  subsequently  used  in  two  of  the  Boston 
schools,  and  with  such  success,  that  the  author  was 
solicited  by  a  number  of  gentlemen,  who  were  ac- 
quainted with  his  "  Elements  of  Algebra,"  and  who 
knew  his  plan  in  the  present  work,  to  prepare  a 
treatise  for  common  schools. 


i\^80f^-^.^5 


VI  PREFACE. 

An  attempt  has  been  made  to  render  the  science 
as  easily  attainable  as  possible,  without  prejudice  to 
the  main  result ;  not  to  save  the  learner  the  trouble 
of  thinking  and  reasoning,  but  to  teach  him  to  think 
and  reason  ;  not  merely  to  supply  a  series  of  simple 
exercises,  but  to  insure  a  good  knowledge  of  the 
subject.  To  what  extent  the  writer  has  attained 
his  object,  is  left  to  intelligent  instructors,  school- 
committees,  and  others,  to  determine. 

Teachers  and  pupils  will  observe  that,  to  repre- 
sent multiplication,  the  full  point  is  generally  used 
in  this  work  rather  than  the  sign  X .  But  to  distin- 
guish the  sign  of  multiplication  from  the  period  used 
as  a  decimal  point,  the  latter  is  elevated  by  inverting 
the  type,  while  the  former  is  larger,  and  placed 
down  even  with  the  lower  extremities  of  the  figures 
or  letters,  between  which  it  stands. 

THOMAS    SHERWIN 


CONTENTS. 


lECT.  PAOS. 

Preliminary  Exercises, 1 

I.    Equations  having  unknown  terms  only  in  the  first 

member,  and  entirely  known  terms  in  the  second,.  .23 
II.     Equations  having  unknown   terms  in  one  member 

only,  and  known  terms  in  both, 27 

III.  Equations  having  both  known  and  unknown  terms  in 

each  member, 34 

IV.  Equations  containing  fractional  parts  of  single  terms, .  .38 
V.     Equations  containing  fractional  parts  of  quantities  con- 
sisting of  several  terms, 42 

Vi.     Equations  which  require  the  subtraction  of  quantities 

containing  negative  terms, 45 

Vit.     Multiplication  of  monomials, 50 

VIII.     Reduction  of  similar  terms, 55 

IX.     Addition, 58 

X.     Subtraction, 60 

XI.     Multiplication  of  polynomials, 64 

XII.     Division  of  monomials, 72 

XIII.  Division  of  polynomials, 74 

XIV.  Multiplication  of  fractions  by  integral  quantities, 83 

XV.     Division  of  fractions  by  integral  quantities 88 

XVI.     Factors,  or  divisors  of  algebraic  quantities, 90 

XVII.     Simplification  of  fractions, 93 

XVIII.     Multiplication  effractions  by  fractions, 96 

XIX.     Least  common  multiple, 97 

XX.     Addition  and  subtraction  of  fractions ;   common  de- 
nominator,   99 

XXI.     Division  of  integral  and  fractional  quantities  by  frac- 
tions,   103 

XXII.     Equations  containing  two  unknown  quantities, 106 

XXIII.     Equatio-"  \g  three  unknown  quantities,  ..  .,119 


¥111 


CONTENTS. 


SECT. 

XXIV. 

XXV. 

XXVI. 

XXVII. 

XXVIII. 

XXIX. 

XXX. 
XXXI. 

XXXII. 

XXXIII. 

XXXIV. 
XXXV. 

XXXVI. 
XXXVII. 
XXXVIII. 

XXXIX. 

XL. 

XLI. 

XLII. 
XLIII. 
XLIV. 

XLV 


PAGE. 

Substitution  of  numbers  in  algebraic  quantities,..  126 

Literal  equations, 129 

Generalization, 130 

Extraction  of  the  second  roots  of  numbers, 142 

Second  roots  of  fractions,  and  extraction  of  second 

roots  by  approximation, 153 

Questions  producing  pure  equations  of  the  second 

degree, 156 

Affected  equations  of  the  second  degree, 157 

Extraction  of  the  third  roots  of  numbers, 167 

Third  roots  effractions,  and  the  extraction  of  third 

roots  by  approximation, 177 

Questions  producing  pure  equations  of  the  third 

degree, 181 

Powers  of  monomials, • 183 

Powers  of  polynomials, 185 

Roots  of  monomials, 192 

Second  roots  of  polynomials, 195 

Transformation   and   simplification  of  irrational 

quantities, 199 

Operations  on  irrational  quantities  with  fractional 

exponents, 205 

Operations  on  irrational  quantities  with  the  rad- 
ical sign, 210 

Ratio  and  proportion, 217 

Progression  by  difference, 228 

Examples  in  progression  by  difierence, 231 

Progression  by  quotient, 232 

Examples  in  progression  by  quotient, •  • .  .237 


COMMO]N[,.,^lgJ^m)t''rtGEBRA. 


PRELIMINARY    EXERCISES. 

Art.  1.  Algebra  has  been  called  **  Universal  Arithme- 
tic," and  is  principally  distinguished  from  common  arith- 
metic by  this,  that  in  algebra  calculations  are  performed 
by  means  of  letters  and  signs ;  letters  being  used  to  rep- 
resent quantities,  and  signs  to  indicate  operations,  or  to 
stand  for  certain  words. 

The  sign  -|-,  called  plus,  which  signifies  more,  repre- 
sents addition.  Thus,  6  -(-  4  represents  the  addition  of  6 
and  4,  or  indicates  the  sum  of  these  numbers. 

The  sign  — ,  called  minus,  which  signifies  less,  repre- 
sents subtraction,  and  is  placed  immediately  before  the 
quantity  to  be  subtracted.  Thus,  8  —  3  represents  the 
subtraction  of  3  from  8,  or  indicates  the  difference  of 
these  numbers. 

Multiplication  is  represented  by  a  full  point,  or  the  sign 
X,  placed  between  the  quantities  to  be  multiplied.  Thus, 
7.3,  or  7  X  3,  represents  the  multiplication  of  7  by  3,  or 
indicates  the  product  of  these  numbers. 

Division  is  repre«<ented  in  the  form  of  a  fraction,  the 
dividend  being  placed  over  the  divisor ;  also  by  the  sign  : 
or  -r-.  Thus,  each  of  the  expressions,  ^,3:7,  and 
3  —  7,  represents  the  division  of  3  by  7,  or  indicates  the 
Quotient  arising  from  that  division  When  the  sign  :  oi 
1 


'4  PRELIMINARY    EXERCISES. 

-|-  is  used,  the  quantity  preceding  the  sign  is  the  dividend^ 
and  that  following  it,  the  divisor. 

Instead  of  the  words  "  equal  to/'  **  equals/'  or  others 
of  similar  meaning,  we  use  the  sign  =,  which  is  called 
the  sign  of  equality.  Thus,  5  -f-  3zi:  8.  is  read,  *'  5  plus 
3  equals  8." 

Instead  of  the  words  **  greater  than,"  or  "  less  than,"  we 
use  the  sign  ^  or  <^,  called  the  sign  ot  inequality.  Thus, 
6  >  4  is  read,  "  (3  greater  than  4/'  and  3  <^  5  is  read, 
**  3  less  than  5  ;  "  the  open  end  of  the  sign  being  turned 
towards   the  greater   quantity. 

The  sign  .-.  is  used  instead  of  the  word  ''therefore" 
or  *'  consequently." 

Art.  2.  An  axiom  is  a  self-evident  truth ;  the  follow 
ing  are  of  this  nature. 

Axioms. 

1.  If  the  same  quantity,  or  equal  quantities,  be  added 
to  equal  quantities,  the  sums  will  be  equal. 

2.  If  the  same  quantity,  or  equal  quantities,  be  suh' 
tr acted    from    ecjual     quantities,    the    remainders    will    be 

.  '^qual. 

3.  If  equal  quantities  be  rnultiplicd  by  the  same  quantity, 
or  by  equal  quantities,  ih^  products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  the  same  quantity, 
or  by  equal  quantities,  the  quotients  will  be  equal. 

5.  If  the  same  quantity  be  both  added  to  and  subtracted 
from  another,  the  value  of  the  latter  will  not  be  changed. 

0.    If  a  quantity  he  both  multiplied  and  divided  by  anoth- 
ic    .v  ■{]  not  be  changed. 

7.  Two  (juantities,  each  of  which  is  equal  to  a  thirds 
are  equal  to  e»ch  other. 

8.  The  w*.  ie  of  a  quantity  is  greater  than  a  part  of  it 


PRELIMINARY    EXERCISES.  3 

9.  The  whole  of  a  quantity  is  equal  to  the  sun  of  all 
its  parts. 

Art.  3.  I.  Two  boys,  A  and  B,  had  together  40 
apples ;  but  B  had  3  times  as  many  as  A.  How  many 
had  each? 

It  is  manifest  that,  if  we  knew  the  number  A  had,  by 
multiplying  by  3  we  should  find  the  number  B  had.  The 
number  that  6  had  may,  therefore,  be  called  the  unknown 
quantity.  From  the  conditions  of  the  question,  we  know 
that  once  the  unknown  quantity  and  3  times  that  quantity 
must  make  40.  But  for  conciseness,  and  in  order  to  avoid 
the  repetition  of  the  words  "  unknown  quantity,"  we  may 
use  a  symbol  to  represent  that  quantity.  The  symbols 
used  in  algebra  to  represent  unknown  quantities,  are  the 
last  letters  of  the  alphabet,  as  x,  y,  z;  the  first  letters 
of  the  alphabet,  when  used,  commonly  represent  known 
quantities. 

Let  X  represent,  in  the  question,  the  number  of  apples 
A  had.  Then  3  x  will  represent  the  number  B  had. 
Hence, 

X'\-^  X  ^=.  40.     Combining  x  and  3  x,  we  have 
4  X  =  40  ;  and  1  2;  or  x  must  be  J  of  40,  or 

X  =  10;  the  number  A  had , 
^xzzz  30,  the  number  B  had. 
2.  The  united    ages  of  three  brothers.  A,  B,   and  C, 
amount  to   120  years.     B  is  twice  as  old  as  A,  and  C's 
age  is  equal  to  the  sum  of  the  ages  of  his  two  brothers. 
Required  the  age  of  each. 

Let  X  represent  A*s  age ;  then  B's  will  be  represented 
by  2  X,  and  C's  hy  x -\-^x,  ox  ^x.     Hence, 

x-\-2x-\-Zx^:z  120.     Combining,  we  have 
6  x  =  120  ; 
.•.  X  =  20  years,  A's  age  * 


4  PRELIMINARY    EXERCISES. 

2  X  1=  40  years,  B's  age;  and 

3  x  =  60  years,  C's  age. 

We  have  seen,  in  the  two  preceding  questions,  the  use 
which  may  be  made  of  letters  to  represent  quantities  whose 
value  is  to  be  found.  We  shall  now  give  some  exercises 
in  representing  quantities,  and  in  performing  very  simple 
operations. 

Art.  4:.  1.  If  a:  represent  the  price  of  an  apple  in 
cents,  what  will  represent  the  price  of  2  apples  ?  Of  3, 
4,  5,  6  ?  Ans.  2  x,  3  x,  4  x,  5  x,  6  x. 

2.  If  one  pear  cost  x  cents,  what  will  4,  6,  9,  11  pears 
cost? 

3.  If  X  represent  the  number  of  miles  a  man  can  travel 
in  one  day,  what  will  represent  the  number  of  miles  he 
can  travel  in  3,  5,  7,  10,  15,  30  days? 

4.  If  one  yard  of  cloth  cost  x  dollars,  what  will  repre- 
sent the  price  of  2,  7,  9,  18,  20  yards  ? 

5.  If  2  X  dollars  represent  the  price  of  one  barrel  of 
flour,  what  will  represent  the  price  of  2,  3,  4,  5,  6,  7, 
8,  9,  10  barrels? 

6.  If  3  X  represent  the  number  of  shillings  a  man  earns 
in  a  day,  what  will  represent  the  number  of  shillings  he 
earns  in  2,  5,  6,  8,  10,  12  days  ? 

7.  If  4  x  cents  represent  the  value  of  one  pound  of 
coffee,  what  will  represent  the  value  of  3,  6,  12,  13 
pounds  ? 

8.  If  5  X  represent  the  weight  of  one  bag  of  coffee  in 
pounds,  what  will  represent  the  weight  of  2,  3,  4,  7,  9, 
15  bags  ? 

Remark.  The  learner  will  perceive  that  the  preceding 
are  examples  of  multiplication.  Thus,  3  x  is  the  product 
of  3  and  x,  and  12  x  is  the  product  of  x  and  12,  of  2  a 
ajod  0,  or  of  3  x  and  4. 


PRELIMINARY    EXERCISES.  ,  0 

9.  How  much  is  5  times  xl  Ans.  5  z.  How  much  ib 
X  times  5  ?     Ans.  5  x. 

10.  If  one  sheep  cost  9  s.,  what  will  represent  the 
ptice  of  a  number  x  of  sheep  ?     Of  2  x,   3  x,  4  a:,  5  x, 

6  X  sheep  ? 

11.  If  one  man  earn  10  s.  per  day,  what  will  represent 
the  daily  wages  of  a:,  2  x,  3  x,  5  a:,  7  a:,  11a:  men  ? 

12.  If  X  represent  the  daily  wages  of  one  man  in  shil- 
lings, what  will  represent  the  earnings  of  2  men  in  3 
days,  of  3  men  in  3  days  ?  of  4,  5,  6  men  in  the  same 
time? 

13.  If  X  represent  the  daily  wages  of  cne  man,  what 
will  represent  the  wages  of  2  men  for  5,  6,  7,  8,  9,  10 
days? 

14.  If  X  represent  the  daily  wages  of  one  man,  what 
will  represent  the  wages  of  2  men  for  3  days  ?  of  3  men 
for  4  days?  of  4  men  for  5  days?  of  6  men  for  7  days?  of 

7  men  for  8  days  ? 

15.  If  X  represent,  in  shillings,  the  price  of  one  yard 
of  cloth,  what  will  be  the  price  of  1  piece,  3  yds.  long  ?  of 
2  pieces,  each  4  yds.  long?  of  5  pieces,  each  6  yds.  long? 

16.  If  one  man  earn  $2  per  day,  what  will  represent 
(he  wages  of  2,  3,  4,  5,  6,  7  men  for  3  a;  days?  And 
what  will  r^resent  the  wages  of  3,  5,  7,  11,  13  men  for 

5  X  days  ? 

17.  What  is  7  times  2  a:?  8  times  3  a:?  10  times  5x1 
12  times  6 a:? 

18.  What  is  x  times  4  s.?     2  x  times  5  s.?     3  a:  times 

6  s.?     7  a:  times  9  s.  ? 

19.  If  X  represent  the  price  of  a  cow,  and  an  ox  is 
worth  twice  as  much  as  a  cow,  what  will  represent  the 
price  of  5  oxen  ? 

20.  If  a  yard  of  black  cloth  cost  twice  as  much  as  a 
yd.  of  white,  and  a  yd.  of  blue  cost  3  times  as  much  as  a 


O  PRELIMINARY    EXERCISES. 

yd.  of  black,  what  will  represent  the  price  of  4  yds.  of 
black,  and  what  the  price  of  5  yds.  of  blue,  the  price  of 
a  yd.  of  white  being  represented  by  x? 

AuT.  5.     The  addition  of  quantities,  as  has  been  al- 
ready   stated,    is   expressed    by  means   of  the    sign  -f- 
Thus,  3  -f-  4  -f-  6,  means  that  3,  4,  and  6  are  added  to- 
gether, and  is  read,  *^  3  plus  4  plus  6." 

1.  If  x  represent  the  price  of  an  apple,  and  2  x  that  of 
a  pear,  what  will  represent  the  price  of  an  apple  and  a 
pear  together  ?  Ans.  x  -\-2x, 

2.  If  3  X  represent  the  price  of  a  cow,  and  4  x  that  of 
an  ox,  what  will  represent  the  price  of  both  ? 

3.  If  X,  2  X,  and  3  x  represent  the  respective  ages  of  3 
men,  what  will  represent  the  sum  of  their  ages  ? 

4.  If  3  a:,  4  x,  5  x,  and  6  x  represent  the  respective 
lengths  of  4  pieces  of  cloth,  what  will  represent  the 
length  of  the  whole  ? 

5.  If  apples  cost  2  cents  each,  and  pears  3  cents  each, 
what  will  represent  the  whole  cost  of  a:  apples  and  x  pears? 

6.  If  corn  cost  4  s.,  rye  6  s.,  and  wheat  8  s.  per  bushel, 
what  will  represent  the  sum  of  the  prices  of  x  bushels  of 
each? 

7.  What  will  represent  the  entire  price  o?  3  x  bushels 
of  rye,  at  5  s.  a  bushel,  and  6  x  bushels  of  wheat,  at  9  s.  a 
bushel  ? 

8.  If  X  represent  A's  money,  and  B  have  3  times  and  C 
4  times  as  much  as  A,  what  will  represent  the  amount  of 
their  money  ? 

9.  If  X,  2  X,  and  3  x  represent  the  respective  daily  wages 
of  A,  B,  and  C,  what  will  represent  the  amount  of  their 
wages,  A  working  9,  B  6,  and  C  5  days  ? 

10.  A  man  has  3  sons,  whose  ages  are  such  that  the 
2d  is  twice  as  old  as  the  youngest,  and  the  eldest  twice  as 


PRELIMINARY    EXERCISES.  1 

old  as  the  2d.     What  will  represent  the  sum  of  their  ages, 
that  of  the  youngest  being  represented  by  a:  ? 

11.  A  has  twice  as  much  money  as  B,  and  B  3  timeh 
as  much  as  C.  What  will  represent  the  amount  of  their 
money,  if  x  represent  C's  ? 

12.  A  is  of  a  certain  age,  represented  by  r,  B  twice  aa 
old,  C  3  times  as  old  as  B,  and  D  as  old  as  B  and  C  both. 
What  will  represent  the  united  ages  of  all  four  ? 

13.  If  John's  money  be  represented  by  a:,  and  Joseph's 
by  y,  what  will  represent  the  amount  of  their  money  1 

14.  If  X  represent  A's  age,  and  y  B's,  what  will  repre- 
.'Sent  the  sum  of  3  times  A's  and  7  times  B's? 

15.  If  3  X  represent  the  price  of  a  barrel  of  beer,  and 
2  y  the  price  of  a  barrel  of  cider,  what  will  represent  the 
whole  cost  of  7  barrels  of  beer  and  5  of  cider  ? 

16.  If  X  represent  A's  age,  and  y  B's,  and  C  be  twice  as 
old  as  A,  and  D  3  times  as  old  as  B,  what  will  represent 
the  united  ages  of  all  1 

17.  If  one  town  be  3  x  miles  north  of  Boston,  and  an- 
other 4  X  miles  south  of  Boston,  what  will  represent  the 
distance  of  these  towns  asunder  ? 

18.  Two  men  start,  at  the  same  time,  from  the  same 
place,  and  travel  in  opposite  directions.  If  one  travel 
X  miles  and'1:he  other  y  miles  per  hour,  what  will  repre- 
sent their  distance  apart  at  the  end  of  2,  3,  5,  7  hours 
respectively  ? 

Art.  6.  1.  If  4  a:  represent  the  price  of  2  yds.  of 
cloth,  what  will  represent  the  price  of  one  yd.  ? 

Ans,  2  X. 

2.  If  10  X  represent  the  price  of  2  oxen,  what  will 
represei.t  that  of  one  ? 

3.  If  15  X  represent  the  worth  of  5  barrels  of  flour, 
what  will  represent  that  of  1,  2,  3,  4  barrels  ? 

Ans.  Sx,6x,9x,12x 


8  PRELIMINARY    EXERCISES.         * 

4.  If  20  X  represent  the  price  of  10  peaches,  what  uill 
represent  the  price  of  1,  2,  3,  4,  5,  6,  7,  8,  9  peaches  ? 

5.  When  12  x  represents  the  worth  of  4  yds.  of  cloth, 
what  will  represent  that  of  1,  2,  3,  5,  6,  7,  8,  9,  10, 
12  yds.? 

G.  When  9  x  represents  the  number  of  miles  a  man 
travels  in  3  days,  what  will  represent  the  number  of  miles 
he  travels  in  5,  7,  9,  11,  20  days  ? 

7.  If  6  a:  represent  the  price  of  3  eggs,  what  will  repre- 
sent that  of  9,  12,  13,  15,  17  eggs  ? 

8.  If  27  X  represent  the  number  of  miles  a  ship  sails  iu 

9  hours,  what  will   represent  the  distance  sailed  over  in 
12,  15,  17,  24,  48  hours  ? 

9.  If  48  X  represent  the  distance  a  vessel  sails  in  a  day, 
or  24  hours,  what  will  represent  the  distance  she  sails  in 
1^,2,  2^,3,3-1  days? 

10.  If  4  X  represent  the  price  of  a  bushel  of  corn,  what 
will  represent  that  of  ^  a  bushel?    of  1^,  2J-,  3^  bushels? 

11.  If  2  a:  represent  the  price  of  a  bushel  of  wheat, 
and  wheat  be  worth  twice  as  much  as  corn,  what  will  rep- 
resent the  price  of  3,  5,  7,  9,  10,  12  bushels  of  corn  ? 

12.  If  2  x  represent  the  price  of  corn  per  bushel,  and 
wheat  be  worth  3  times  as  much  as  corn,  and  rye  J  as 
much  as  wheat,  what  will  represent  the  price  of  1,  2,  3, 
4,  5,  6,  7  bushels  of  rye  ? 

13.  If  6  X  represent  the  price  of  a  barrel  of  flour,  and 
a  barrel  of  sugar  be  worth  2  barrels  of  flour,  what  will 
represent  the  price  of  1,  3,  5,  7,  8,  9  barrels  of  rice,  sup- 

^  posing  rice  worth  ^  as  much  per  barrel  as  sugar  ? 
^  14.  If  4  a;  represent  the  price  of  2  barrels  of  cider,  and 
6  X  that  of  3  barrels  of  beer,  what  will  represent  the  sum 
of  the  prices  of  1  barrel  of  each?  of  4  barrels  of  beer 
and  2  of  cider  ?  of  6  barrels  of  beer  and  5  of  cider '? 
of  9  barrels  of  beer  and  14  of  cider  '? 


^ 


PRELIMINARY    EXERCISES.  9 

15.  If  10  X  represent  the  weight  of  5  boxes  of  sugar, 
and  4  x  that  of  2  boxes  of  raisins,  what  will  represent  the 
whole  weight  of  3  boxes  of  sugar  and  5  of  raisins  ?  8  of 
sugar  and  7  of  raisins  ?     17  of  sugar  and  20  of  raisins  ? 

16.  When  4  x  represents  the  price  of  corn,  and  8  x  the 
price  of  beans,  per  bushel,  what  will  represent  the  entire 
price  of  3  pecks  of  corn  and  6  pecks  of  beans  ?  9  pecks 
of  corn  and  7  pecks  of  beans  ? 

17.  Supposing  that  6  x  represents  the  distance  one  man 
travels  in  3  days,  and  8  y  the  distance  another  travels  in  4 
days ;  if  they  start  at  the  same  time,  from  the  same  place, 
and  travel  in  opposite  directions,  what  will  represent  their 
distance  apart  at  the  end  of  1  day  ?    of  2,  3,  4,  5  days? 


Art.  7.  1.  If  x  represent  A's  age,  and  B  be  2  years 
older,  what  will  represent  B's  age  ?  Arts,  x  -f-  2. 

2.  If  X  represent  A's  age  in  years,  and  B  be  9  months 
older,  what  will  represent  his  age  in  months  ? 

Ans,  12  a: +  9. 

3.  If  a:  represent  the  distance,  in  miles,  of  a  certain 
town  from  N.  York,  what  will  represent  the  distance  from 
that  city  of  another  town,  12  miles  more  remote  ? 

4.  If  2  a;  shillings  be  the  price  of  corn  per  bushel,  and 
wheat  be  worth  4  s.  per  bushel  more  than  corn,  what  will 
represent  the  price  of  a  bushel  of  wheat  ? 

5.  If  X  represent  A's  age  in  years,  what  will  represent 
B's  age,  and  what  C's  ;  B  being  2  years  older  than  A,  and 
C  3  years  older  than  B  ? 

Ans,  B's,  X  -f  2  ;   and  C's,  x  +  2  +  3,  or  x  +  5. 

6.  A  man  has  4  sons,  each  of  whom  is  3  years  older 
than  his  next  younger  brother.  What  will  represent  the 
ages  of  the  3  elder,  if  x  represent  the  age  of  the  youngest 
in  years  ? 

7.  If  black  cloth  cost  4  s   a  yard  more  than  white,  and 


10  PRELIMINARY    EXERCISES. 

blue  5  s.  a  yard  more  than  black,  what  wih  represent  the 
price  of  a  yard  of  the  black  and  blue  respectively,  x 
representing  the  number  of  shillings  given  for  a  yard  of 
white  ? 

8.  A  started  from  Boston  2  hours  before  B,  and  3  hours 
before  C,  and  they  all  arrived  at  Philadelphia  at  the  same 
time.  What  will  represent  the  time  B  and  C  respectively 
were  on  the  road,  x  representing  the  number  of  hours  in 
which  A  performed  the  journey  ? 

9.  A's  horse  is  worth  $5  more  than  B's,  and  B's  is 
worth  one  eagle  more  than  C's.  What  will  represent,  in 
dollars,  the  price  of  A's  and  B's  respectively,  if  x  repre- 
sent that  of  C's  ? 

10.  A  is  worth  $100  more  than  twice  what  B  is  worth. 
What  will  represent  A's  estate,  if  x  represent  B's,  in  dollars  ? 

11.  A  is  4  years  more  than  twice  as  old  as  B,  and  B  is 
3  times  as  old  as  C.  What  will  represent  the  age  of  A  and 
B  respectively,  if  x  represent  the  number  of  years  in  C's  ? 

12.  A  can  earn  ^  as  much  as  B,  and  C  2  s.  more  than 
twice  as  much  as  B,  in  a  day.  What  will  represent  B's 
and  C's  daily  wages,  if  x  represent  the  number  of  shillings 
A  earns  per  day  ? 

13.  A  man  has  a  number  x  of  swine,  5  more  than  3 
times  as  many  cows  as  swine ;  and  his  number  of  sheep  is 
equal  to  that  of  his  swine  and  cows  together.  What  will 
represent  the  number  of  cows  and  sheep  respectively  1 
and  what  will  represent  the  aggregate  number  of  his 
whole  stock  ? 

14.  What  will  represent  4  more  than  twice  Sxl  7 
more  than  4  times  3  2:?     10  more  than  3  times  5x1 

15.  What  will  represent  3  more  than  ^  of  4  x  ?  10 
more  than  ^  of  9  x  ?     7  more  than  -J  of  20  a:  ? 

16.  What  will  represent  the  sum  of  x  and  2  ?  of  3  a 
and  4  ?    of  6  x  and  9  ?    of  7  x  and  10  I 


PRELIMINARY    EXERCISES.  li 

17.  A  is  2  years  older  than  B,  and  B  4  years  older  than 
C.  What  will  represent  the  sum  of  their  ages,  C's  being 
represented  by  a:  ? 

18.  B  has  $4  more  than  A,  and  $5  less  than  C.  What 
will  represent  the  number  of  dollars  B  and  C  have,  respec- 
tively, that  which  A  has  being  x1 

19.  A  begins  trade  with  x  dollars,  and  gains  $100;  B 
l;5gins  with  twice  as  much  money  as  A,  and  gains  $150; 
C  begins  with  as  much  as  A  and  B  both,  and  gains  $70. 
What  will  represent  the  amount  of  their  property,  includ- 
ing stock  and  gain  ? 

> 

Art.  8.  1.  If  x  represent  the  number  of  dollars  A 
has,  and  B  have  $2  less,  what  will  represent  B's  money? 

Ans.  X  —  2. 

2.  If  X  represent  the  number  of  cows  a  ftirmer  has, 
what  will  represent  the  number  of  his  cows  after  he  shall 
have  sold  10? 

3.  If  X  shillings  represent  the  price  of  a  bushel  of  wheat, 
and  rye  be  4  s.  a  bushel  cheaper,  what  will  represent  the 
price  of  a  bushel  of  rye  ? 

4.  A  drover,  having  100  sheep,  sold  a  number  of  them 
represented  by  x.  What  will  represent  the  number  he  had 
remaining?  Ans.  100 — x, 

5.  Of  150  acres  of  land,  a  part  is  cultivated,  and  the 
rest  is  woodland.  If  x  represent  the  number  of  acres 
cultivated,  what  will  represent  the  iiumber  of  acres  of 
woodland? 

6.  The  sum  of  two  numbers  being  20,  what  will  repre- 
sent the  second,  if  x  represent  the  first?- 

7.  A  man  has  59  coins,  consisting  of  dollars  and  sove- 
reigns. If  X  represent  the  number  of  dollars,  what  will 
represent  that  of  the  sovereigns? 

8.  A  walked  a  number  of  miles,   represented  by  x ;  B 


12  PRELIMINARY    FAERCISES. 

walked  twice  as  far  wanting  4  miles.    What  will  represent 
the  distance  B  walked  ? 

9.  A  man  having  a  certain  number,  x,  of  dollars, 
doubled  his  money,  and  afterwards  lost  $12.  What  will 
represent  the  money  he  then  had  ? 

10.  A  draper,  having  50  yards  of  cloth,  sold  a  num- 
ber of  yards,  represented  by  x.  What  will  represent  the 
number  of  yards  remaining? 

11.  A  grocer,  having  500  lbs.  of  coffee,  sold  3  bags, 
each  containing  x  lbs.  What  will  represent  the  number 
of  pounds  remaining  ? 

12.  A  has  X,  and  B,  y  dollars.  If  A  give  B  f  5,  what 
will  represent  the  number  of  dollars  each  has  then  ? 

Ans.  A,  X  —  5,  and  B,  y-^S, 

13.  If  A  have  3  x,  and  B,  2  y  dollars,  what  will  represent 
their  money,  after  B  shall  have  given  A  $10? 

14.  A  farmer  has  100  sheep  in  one  pasture  and  75  in 
another.  What  will  represent  the  number  in  each  flock, 
after  he  shall  have  taken  2  x  sheep  from  the  larger  and 
put  them   with  the  smaller 'i! 

15.  A  and  B  have  each  50  cows.  A  sells  a  number,  x, 
and  B  sells  3  times  as  many.  What  will  represent  the 
number  each  has  left  ? 

\Q.  A  poulterer  has  x  turkeys  and  y  geese.  After  sell- 
ing 4  turkeys  for  15  geese,  what  will  represent  the  number 
of  each  he  then  has  ? 

17.  If  a  woman  be  x  years  old,  and  her  husband  twice 
as  old  as  she,  at  the  time  of  marriage,  what  will  represent 
their  respective  ages  5  years  before  marriage? 

K 

Art.  9.  According  to  Ax.  5,  if  the  same  quantity  be 
both  added  to,  and  subtracted  from  another,  that  other 
will  not  be  changed  in  value.     Thus,  if  2  be  added  to  8, 


PRELIMINARY    EXERCISES.  13 

the  sum  will  be  10.  Now,  if  2  be  subtracted  from  10, 
the  remainder  will  be  8,  the  same  as  at  first. 

If  these  operations  are  represented  merely,  the  result 
'  .  will  be  expressed  thus,  8  -|-  2  —  2,  in  which  -f-  2  and  — 2 
destroy  or  cancel  each  other,  and  leave  8. 

But  if  more  is  added  to  a  quantity  than  is  subtracted 
from  it,  that  quantity  is  increased  by  the  excess  of  what  is 
added  above  what  is  subtracted.  Thus,  12  +  7  —  6  is  the 
same  as  12  -f- 1,  or  13,  because  1  more  is  added  to  12  than 
is  subtracted  from  it. 

On  the  other  hand,  if  more  is  subtracted  from  a  quan- 
tity than  is  added  to  it,  that  quantity  is  diminished  by  the 
excess  of  what  is  subtracted  above  what  is  added.  Thus, 
15  —  7  +  3,  or  15  +  3  —  7,  is  the  same  as  15  —  4,  or  11. 

In  like  manner,  x  +  5  —  3,  is  the  same  as  x  +  2  ;  and 
X  —  8  +  5,  is  the  same  as  x  —  3. 

1.  What  is  the  same  as  z  —  4  +  4?  asx  +  6  —  6?  as 
^  — 5-^x1    as— 7  +  7  +  2x7 

2.  What  is  the  same  as  X  +  9  —  3?  as  x  —  9  +  3?  as 
2x  — 7  +  3?  as3  — 7  +  2a:?  as5x  — 5  +  7?  as52;  + 
7  —  9? 

X  3.  What  is  the  same  as4x  — 6  +  3?  as7a:+10  — 4? 
as  12x— 12  +  7? 

The  several  parts  of  an  algebraic  quantity,  connected 
together  by  the  signs  +  and  — ,  are  called  terms.  Thus, 
in  1 X  —  10  +  5  x  +  3,  1 X,  —  10,  5  2:,  and  3  are  the  sep- 
arate terms. 

When  a  quantity  contains  many  terms,  all  those  con- 
sisting of  numbers  only  may  be  reduced  to  one  single 
term,  and  those  which  are  alike  with  regard  to  the  letters, 
to  another.  But  it  must  be  remembered,  that  when  a  term 
has  no  sign  before  it,  it  is  supposed  to  have  the  sign  +. 
Thus,  x  +  2a:  +  52:is  the  same  as8x;4:i  -\- ^ -\- '^  x -{' 1 
is  the  same  as72:  +  9;   10 a:  —  2  +  3z—  5   is  the  same 


14  PRELIMINARY    EXERCISES. 

as  13  X  — -  7.  In  the  last  example,  10  x  and  3  x  are  added, 
whilst  2  and  5  are  both  subtracted,  which  is  the  same  as 
subtracting  their  sum,  7 

In  a  similar  manner,  10  2:-[-7-|-3x  —  5  is  the  same  . 
as   13a:  +  2;    12x  — 8  +  42;  +  3  is  16a:  — 5;   and  12x 
j-3_3a:_8is9x  —  5. 

Again,4x  +  10  +  3  2:  — 7  +  8a:  — 3— 4x  +  6  — 3x 
-4is82:  +  2;  3  x  — 7 +  4x  + 2  — 3:+ 1 +62:  — 3  — 
5a:is  62^  —  7;   12  — 3a:  +  7-f2x  — 3  — 5x  is  16  — 
6x;  and  7  +  32:  — 4  +  2x  —  x-|-2  is  5  +  4a:. 

In  the  last  four  examples,  we  combine  all  the  terms  con- 
taining X,  and  preceded  by  the  sign  -f-?  expressed  or  im- 
plied ;  then  combine  all  those  containing  x,  and  preceded 
by  the  sign  — ,  and  take  the  difference  of  the  two  sums, 
giving  it  the  sign  of  the  greater  sum.  With  the  numbers 
we  proceed  in  the  same  way. 

4.  What  is  the  same  as  x-|- 2  4-3  x  —  4-f-5  2:-f-7?  as 
4a;  — 2  +  2X  +  7  +  32:  — 9  +  x-f-lO?  asllz  — 6  + 
3x  +  7  — 6x  — 4?  as  7  +  3x  — 6  +  4x4-10  — 13x? 
as5x4-7— lOx  +  3  — 4x  — 24-X4-5? 

Let  the  answers  to  the  following  questions  be  reduced 
in  the  manner  shown  above. 

5.  A  is  X  years  old ;  B  is  twice  as  old  as  A,  and  three 
years  more ;  and  C's  age  is  2  years  less  than  the  sum  of 
A's  and  B's.     What  will  represent  C's  age  ? 

Ans.  x4-2x4-3  — 2,  or  3x4-1. 

6.  A  has  X  dollars,  and  B  has  twice  as  many.  A  gains 
$20,  and  B  loses  $5.  What  will  then  represent  the  sum 
of  A's  and  B's  money  ? 

Ans.   x4-204"2x  — 5,  or  3x4-15. 

7.  A  has  $50,  and  B  $30.  A  loses  x  dollars,  and  B 
gains  4  times  as  much  money  as  A  loses.  What  will  then 
represent  the  amount  of  their  money  ? 

Ans.  50  — x4-30  4-4x,  or  80  4-3x. 


</ "-^-CC^^-t^^UU: 


PRELIMINARY    EXERCISES.  19 

Observe  that,  in  the  three  preceding  examples,  the  addi- 
tion of  quantities  containing  several  terms  is  performed 
by  v^riting  them  after  each  other,  without  altering  the 
signs ;  after  which  the  result  is  reduced.  That  the  signs 
should  not  be  changed,  may  easily  be  shown  by  figures. 
Thus,  12  —  5  and  17  —  3,  when  added,  give  12  —  5  + 
17  —  3,  which  is  the  same  as  29  —  8  or  21  ;  for  12  —  5 
i.r  7,  and  17  —  3  is  14,  and  the  sum  of  7  and  14  is  21. 

8.  An  army  consists  of  x  offi.cers,  6  more  than  3  times 
as  many  cavalry,  and  30  less  than  10  times  as  many  in- 
fantry as  officers.  What  will  represent  the  number  of  men 
in  the  army  ? 

9.  A  is  10  years  younger,  and  C  20  years  older,  than 
B.  What  will  represent  the  sum  of  their  ages,  if  x  rep- 
resent the  number  of  years  in  B's  age  ? 

10.  Four  towns  are  in  a  straight  line,  and  in  the  order 
of  the  letters  A,  B,  C,  and  D.  The  distance  from  A  to 
B  is  20  miles  more,  and  the  distance  from  C  to  D  is  30 
miles  less,  than  the  distance  from  B  to  C.  What  will 
represent  the  whole  distance  from  A  to  D,  if  x  represent 
the  number  of  miles  from  B  to  C  ? 

11.  Three  men  owed  x  guineas.  A  could  pay  the  whole 
debt  wanting  10  guineas,  B  could  pay  it  wanting  20  guin- 
eas, and  C  could  pay  it  and  have  15  guineas  left.  What 
will  represent  the  number  of  guineas  they  all  had  ? 

12.  A  man  has  lived  x  years  in  France,  5  years  more 
than  twice  as  long  in  England,  and  15  years  less  than  3 
times  as  long  in  America.  These  being  his  only  places 
of  residence,  what  will  represent  his  age? 

Art.  10.  When  a  quantity  containing  several  terms 
is  to  be  multiplied,  each  term  must  be  multiplied,  and  tne 
signs  remain  unchanged,  except  under  particular  circum* 
stances,  which  will  be  hereafter  explained. 


16  PRELIMINARY    EXERCISES. 

Thus,  3  times  2  +  5  is  6  +  15;  for  2  +  5  is  7,  and  3 
times  7  is  21,  the  same  as  6+15.  In  like  manner,  5 
times  a:  +  10  is  5  x  +  50. 

Again,  4  times  9  —  3  is  36  —  12  ;  for  9  —  3  is  6,  and 
4  times  6  is  24,  the  same  as  36  —  12.  In  like  manner,  7 
times  3  a:  —  4  is  21  x  —  28,  and  9  times  7  —  2  a;  is  63 — ■ 
18  X. 

1.  A,  having  x  dollars,  gains  $3,  and  afterwards  doubles 
his  money.     What  will  represent  what  he  then  has? 

Ans.  2  a: +  6. 

2.  A  man,  having  3  x  sheep,  sells  5  of  them,  and  after- 
wards triples  his  stock.  What  will  then  represent  the 
number  of  his  sheep  ?  Ans.  9x  —  15. 

3.  The  ages  of  three  brothers  are  as  follows :  the  age 
of  the  eldest  is  4  years  more  than  twice  that  of  the  sec- 
ond, and  that  of  the  second  is  2  years  more  than  3 
times  that  of  the  youngest.  What  will  represent  the  sum 
of  their  ages,  the  youngest  being  x  years  old  ? 

4.  Two  men,  A  and  B,  have  together  $100.  At  the 
end  of  a  year,  A  has  $5  more  than  double  what  he  had 
at  first,  and  B  has  $20  less  than  3  times  what  he  had  at 
first.  What  will  represent  the  amount  of  what  both  had 
then,  if  x  represent  the  n^umber  of  dollars  A  had  at  first? 

5.  What  is  5  times  the  quantity  represented  by  8  a: 
+  5?    7  times  20  — 3x?    9  times  3a:  +  y  —  5? 

6.  A  has  X  dollars,  and  B  has  $5  more  than  twice  as 
much.  B  gives  A  $7,  after  which  each  doubles  the 
money  he  then  has.  What  will  then  represent  the  money 
each  has  ?  what  the  amount  of  their  money  ? 

7.  A  lends  B  $5,  and  doubles  the  money  he  has  left ; 
after  which  B  doubles  his  money,  and  then  repays  A. 
What  will  then  represent  the  stock  of  each,  and  what  their 
joint  stock,  if  A  had  4  a;  dollars  at  first,  and  B  half  as 
much? 


PRELIMINARY    EXERCISES.  11 

8.  A  man  paid  away  $3  each  morning,  and  douoled 
the  remainder  of  his  money  during  the  day.  If  x  repre- 
sent the  number  of  dollars  he  had  at  first,  what  will  rep- 
resent his  money  at  the  end  of  the  5th  day  ? 

9.  A  merchant  doubles  his  stock  each  year,  wanting 
:£1,000.  What  will  represent  his  stock  at  the  end  of  the 
4th  year,  if  he  begins  trade  with  x  pounds  1 

10.  A  has  <£100,  and  B  <£75.  A  gives  B  x  pounds  of 
his  money,  after  which  A  doubles  his  money,  and  B  triple? 
his.     What  will  then  represent  the  whole  of  their  money  \ 

Art.  11,  When  a  quantity  containing  several  terms 
is  to  be  divided,  each  term  must  be  divided,  and  the  signs 
remain  unchanged,  except  under  particular  circumstances, 
to  be  hereafter  explained. 

Thus,  half  of  10  +  4  is  5  +  2;  for  10  +  4  is  14,  half 
of  which  is  7,  the  same  as  5  +  2.  Again,  ^  of  15  —  6  is 
5  —  2;  for  15  —  6  is  9,  ^  of  which  is  3,  the  same  as 
5  —  2. 

In  like  manner, -i  of  20  +  10  x  is  4  +  2x;  |  of  21  x 
—  14  is  32;  — 2;  ^  of  27  —  182:  is  3  —  22;;  and  i  of 
100  — 20 2;  is  25  — 52;. 

1.  B  is  4  years  older  than  A,  and  C's  age  is  half  of  the 
sum  of  A's  and  B's.  If  x  represent  the  number  of  years 
in  A's  age,  what  will  represent  C's  ?    x 

2.  B  is  4  years  .older  than  A,  and  7  years  younger  than 
C,  and  they  have  a  sister  whose  age  is  ^  of  the  sum  of 
their  ages.  What  will  represent  the  sister's  age,  if  x  rep- 
resent A's  in  years  ?  What  will  represent  hers,  if  x  rep* 
resent  C's  ? 

3.  A  man,  after  having  doubled  his  money,  lost  $6;  he 
afterwards  gained  half  as  much  as  he  then  had.  What 
will  represent  his  last  gain,  if  x  represent  the  number  of 
dollars  he  had  at  first  1 

2 


18  PRELIMINARY    EXERCISES. 

4.  A  and  B  begin  trade  with  equal  stocks.  A  loses 
£50,  and  B  gains  .£100 ;  after  which  each  gains  ^  the 
amount  of  what  both  have.  What  will  represent  what 
each  then  has,  if  each  had  x  pounds  at  first  ? 

5.  A  horse  cost  $12  more  than  3  times  as  much  as  a 
cow,  and  an  ox  f  as  much  as  the  horse.  What  will  rep- 
resent the  price  of  the  ox,  if  the  cow  cost  x  dollars  1 

Ans,  2x  +  8. 

6.  A  man  receives  for  a  year's  wages  12  a:  -|-  60  dollars. 
What  will  represent  his  wages  for  a  month  ?  what  for  7 
months  1 

7.  A  man  paid  for  a  piece  of  blue  cloth  5  s.  more,  and 
for  a  piece  of  white,  20  s.  less,  than  he  did  for  a  piece  of 
black.  What  will  represent  the  average  price  per  piece, 
if  a  piece  of  black  cost  x  shillings? 

8.  A  man  has  6  sons,  each  being  6  years  older  than 
his  next  younger  brother.  What  will  represent  their 
average  age,  if  the  youngest  is  x  years  old? 

Art.  12,  When  division  cannot  be  exactly  per- 
formed, it  is  expressed  in  the  form  of  a  fraction.  Thus, 
^  of  2  is  f,  J  of  5  is  f.     In  like  manner,  ^  of  the  quantity 

X  is  — ;  -^  of  2  2;  is  — ,  the  divisor   always  being  placed 

under  the  dividend. 

1.  A  had  a  number  of  dollars,  represented  by  x,  and 
B  had  half  as  much  money  as  A.     What  will  represent 

B's  money  ?  Ans,  --. 

2.  In  a  certain  school,  ^  of  the  scholars  learn  to  read, 
J  learn  arithmetic,  and  ^  learn  algebra.  What  will  rep- 
resent the  number  in  each  of  these  classes  respectively, 
if  X  represent  the  whole  number  of  scholars  ? 


PRELIMINARY    EXERCISES.  19 

*^.  A  farmer  bought  x  sheep,  and  half  &s  many  cows. 
What  will  represent  the  whole  number  of  both? 

Ans.  x-\ . 

Remark,  When  fractions  are  represented  as  added  or 
subtracted,  the  sign  +  or  —  should  be  placed  even  with 
the'line  separating  numerator  and  denominator. 

4.  A  boy,  having  x  cents,  doubled  his  money,  and  then 
lost  ^  of  what  he  had.  What  will  represent  the  number 
of  cents  he  lost  ? 

5.  A's  age  is  5  years  more  than  ^  of  B's.  What  will 
represent  the -sum  of  their  ages,  if  x  represent  B's  age  in 
years  1 

6.  Two  men  engaged  in  trade  with  equal  stocks.  The 
first  gained  $5  more  than  ^  of  his  stock,  and  the  second 
gained  $10  less  than  J  his  stock.  What  will  represent 
the  sum  of  their  gains,  if  the  stock  of  each  was  x  dollars  ? 

7.  A  woman,  at  the  time  of  her  marriage,  was  ^  as  old 
as  her  husband.  The  husband  being  x  years  old  at  the 
time  of  marriage,  what  will  represent  his  wife's  age  5 
years  before  ?  what  10  years  after  1 

Art.  13.  The  division  of  a  quantity  consisting  of 
several  terms,  is  likewise  represented  by  putting  the 
divisor  under  the  whole  of  the  terms.     Thus,  ^-  of  3  -|"  4 

is  -^  ;  -^  of  7  —  3  is  -^^ .     In  like  manner,  ^  of  x  -f-  2 

IS  -— ^—  ;  4  of  2  ^  —  4  IS  . 

1.    B  had  3  sheep  less  than  twice  the  number  A  had, 

and  C  had  \  as  many  as  B      The  number  A  had  being  z, 

what  will  represent  the  number  C  had  ? 

.  So;— 3 

Ans, . 

5 

2    What  wiL  represent  ^  as  much  as  3x  —  4? 


20  PRELIMINARY    EXERCISES. 

3.  What  will  represent  ^  as  much  as  10  —  xl 

4.  What  will  represent  ^  as  much  as  x-^y  —  10  ? 

5.  What  will  represent  -^  as  much  as  a:  —  2^  +  5? 

6.  A  is  5  years  older  than  B.  What  will  represen 
^  of  the  sum  of  their  ages,  B  being  x  years  old  ? 

7.  A  and  B  had  the  same  number  of  eggs.  A  gave  B 
5,  after  which  each  broke  ^  of  what  he  then  had.  What 
will  represent  the  number  each  broke,  if  each  had  x  eggs 
at  first  ? 

8.  A  man,  having  x  dollars,  doubled  his  money,  then 
gave  away  $20,  and  finally  lost  ^  of  what  he  then  had 
What  will  represent  the  number  of  dollars  lost  ? 

9.  If  a  number,  x,  be  increased  by  2,  and  the  sum  be 
divided  by  3,  and  the  quotient  be  increased  by  7,  what 
will  represent  the  result  ?  j       x  +  2  , 

3     "■"    * 

10.  A  man,  having  x  dollars,  gained  $100,  and  then 
paid  away  -^  of  what  he  had,  wanting  $6.  What  will 
represent  the  number  of  dollars  paid  away? 

11.  If  a  number,  represented  by  x,  be  multiplied  by 
3,  and  20  be  added  to  the  product,  this  result,  be  divided 
by  7,  and  the  quotient  be  increased  by  9,  what  will  repre- 
sent the  final  result  ? 

12.  A  had  x  sheep,  B  twice  as  many  and  11  more,  and 
C  35  more  than  -J  as  many  as  B.  What  will  represent  the 
number  C  had  ? 

13.  A  man,  having  x  dollars,  borrowed  $25,  and  then 
lent  if  of  what  he  had,  wanting  $7.  What  will  represent 
the  number  of  dollars  he  lent  ? 

14.  What  is  7  less  than  -^^jof  x-]- 51 

15.  What  is  25  more  than  yV  of  ^  —  20? 


16.    What  is  i  of  2:4-2  added  to  ^  of  x  —  Sl 

+ 
3 


.         074-2    ,    a:  — 3 

Ans,  ^ 


PRELIMINARY    EXERCISES  ^,1 

17.  What  is  ^  of  x-\-i/  added  to  -^  of  x  —  7? 

18.  Double  a  number  represented  by  x,  increase  the 
product  by  4,  divide  this  result  by  7,  and  increase  the 
quotient  by  33.     What  will  represent  the  final  result? 

19.  A  had  3  times  as  much  money  as  B ;  A  lost  $10, 
and  B  $5.  What  will  then  represent  i  of  A's  money 
added  to  -^  of  B's,  if  B  had  x  dollars  at  first? 

20.  A  and  B  had  each  x  dollars.  A  lost  $35,  and  B 
$25 ;  after  which  each  borrowed  ^  as  much  money  as  the 
other  had  left.  What  will  represent  the  number  of  dollars 
each  had  then  ? 

Art.  14.  1.  If  a;  represent  the  age  of  A  in  years, 
and  B  is  ^  as  old,  what  else  will  represent  A's  age,  if  he 

is  20  years  older  than  B  ?  a       ^    i  on 

^^-  2  T- 

2.  There  is  a  fish,  whose  head  is  5  inches  loi^g,  whose 
tail  is  as  long  as  his  head  and  ^  of  the  length  of  his  body, 
and  whose  body  is  as  long  as  his  head  and  tail.  If  x  rep- 
resent the  length  of  the  body,  what  other  expression  will 
also  represent  the  length  of  the  body  ? 

3.  B  is  10  years  older  than  A,  and  10  years  younger 
than  C.  Moreover,  A's  age  is  ^  6f  the  sum  of  B's  and 
C's.  Let  X  years  be  A's  age,  and  find  another  expression 
for  the  same. 

4.  One  half  of  a  school  learn  to  read,  ^  learn  to  write, 
and  the  remainder,  10,  learn  arithmetic.  Let  x  represent 
the  whole  number  of  scholars,  and  find  another  expression 
for  the  same. 

5.  In  a  mixture  of  tin,  copper,  and  zinc,  ^  of  the 
whole  is  tin,  ^  of  the  whole  is  copper,  and  there  are  5  lbs. 
of  zinc.  Let  x  represent  the  number  of  pounds  in  the 
whole,  and  find  another  expression  for  the  same. 

6.  A  man  has  30  coins  in  his  purse,  viz.  eagles,  del* 


22  PRELIMINARY    EXERCISES 

iars,  and  half-dollars.  He  has  twice  as  many  dollars  as 
eagles,  and  10  more  half-dollars  than  dollars.  If  x  repre- 
sent the  number  of  eagles,  what  expression  will  represent 
the  whole  number  of  coins,  and  consequently  be  equal 
to  30? 

7.  B  is  10  years  older  than  A,  and  3  times  A's  age  is 
equal  to  twice  B's.  If  x  represent  A's  age  in  years,  what 
two  expressions  will  be  equal?     Ans.  3a:  and  2x-|-20. 

8.  Two  men  are  of  the  same  age ;  but  if  one  were  18 
years  younger,  and  the  other  10  years  older,  3  times  the 
age  of  the  former  would  be  the  same  as  twice  the  age  of 
the  latter.  What  two  expressions  will  be  equal,  if  x  rep- 
resent the  age  of  each  in  years  ? 

9.  Two  flocks  of  sheep  are  equal  in  numbers;  but  if  20 
sheep  be  transferred  from  one  to  the  other,  one  flock  will 
then  contain  3  times  as  many  as  the  other.  Find  two  ex- 
pressions which  shall  be  equal  after  this  change. 

10.  Half  of  a  man's  life  was  spent  in  Europe,  ^  of 
it  in  Asia,  and  the  remainder,  which  was  10  years,  in 
America.     Find  two  expressions  for  his  age. 

11.  A  woman's  age  is  §  of  her  husband's,  and  the  dif- 
ference of  their  ages  is  9  years.  Find  two  expressions  for 
the  man's  age. 

12.  A  man,  having  60  dollars,  lost  a  certain  number 
of  them,  after  which  he  found  that  3  times  what  he  lost 
was  the  same  as  twice  what  he  had  remaining.  Find  two 
equal  expressions. 

13.  A  farmer  has  twice  as  many  sheep  as  cows;  he 
buys  60  more  sheep  and  5  more  cows,  and  finds  that  he 
has  4  times  as  many  sheep  as  cows.  Find  two  expressiona 
for  the  number  of  his  sheep  after  this  purchase. 


<§>!.]  EQUATIONS    OF    THE    FIRST    DEGREE.  23 


SECTION   I. 

EQUATIONS  OF  THE  FIRST  DEGREE,  HAVING  UNKNOWN 
TERMS  ONLY  IN  THE  FIRST  MEMBER,  AND  ENTIRELY 
KNOWN     TERMS     IN     THE     SECOND. 

Art.  15*  1.  Two  boys  have,  together,  30  cents;  but 
the  elder  has  twice  as  many  cents  as  the  younger.  How 
many  cents  has  each  ? 

Let  X  represent  the  number  of  cents  the  younger  had ; 
then  2  x  will  represent  the  number  the  elder  had. 

Hence,  x-\-2x=zS0.     Reduce  the  terms  containing  x, 
3  a;  =:  30.     Since  3  x  is  equal  to  30,  1  x  or  x  must 
be  -J-  as  much  ;  therefore, 

x=zlO  cents,  the  number  the  younger  had,  and 
2  X  =  20  cents,  the  number  the  elder  had. 
2.    A  farmer  bought  a  cow  and  a  calf  for  $33.    For  the 
cow  he  gave  10  times  as  much  as  for  the  calf     Required 
the  price  of  each. 

Let  X  represent  the  price  of  the  calf  in  dollars ;  then 
10  X  will  represent  that  of  the  cow. 
Hence,  x  -|- 10  x  =:  33.    Reducing  the  terms  containing  x, 
11  X  z=  33;  1  X  or  X  will  be  y^  as  much ;  ,\ 

X  =  $3,  price  of  the  calf,  and 
10  X  zz:  $30,  price  of  the  cow. 

Problems  in  algebra  can  be  proved,  as  well  as  those  in 
arithmetic.  The  proof  of  the  last  consists  in  adding  the 
price  of  the  cow  to  that  of  the  calf,  and  ascertaining  that 
the  sum  is  $33.  Let  the  learner  prove  the  correctness  of 
his  answers  as  he  advances. 

An  equation  is  a  representation  of  the  equality  of  quan- 
tities.    Thus,  X  +  10  ^  =  33  is  an  equati-^n. 


S4  EQUATIONS    OF    THE    FIRST    DEGREE.  [^  L 

A  member  or  side  of  an  equation,  signifies  the  whole  of 
the  quantities  on  the  same  side  of  the  sign  = ;  the  Jirst 
member  being  on  the  left,  and  the  second  member  on  the 
right  hand  side  of  this  sign. 

An  equation  of  the  ^rst  degree  is  one,  in  which,  after 
it  is  freed  from  fractions,  the  unknown  quantities  are 
neither  multiplied  by  themselves  nor  by  each  other. 

Terms  are  the  separate  parts  of  an  algebraic  expression 
affected  by  the  signs  -[-  and  — .  Those  terms  which  have 
no  sign  prefixed  to  them,  are  supposed  to  have  the  sign 
-j-  ;  and  a  term  is  said  to  be  affected  by  a  sign,  when  it  is 
immediately  preceded  by  that  sign,  either  expressed  or 
understood. 

When  the  first  term  of  a  member  of  an  equation,  or  of 
any  algebraic  quantity,  is  affected  by  the  sign  -|-,  it  is 
usual  to  omit  writing  the  sign  before  that  term  ;  but  the 
sign  —  must  always  be  written  before  any  term  affected 
by  it.  Moreover,  terms  affected  by  the  sign  -f-  are  called 
positive ;  those  affected  by  the  sign  —  are  called  negative 
quantities. 

The  equation  a:  -j-  10  z  z=  33  consists  of  three  terms, 
two  in  the  first  member  and  one  in  the  second  ;  and  each 
of  these  terms  is  affected  by  the  sign  -|-,  although  that 
sign  is  written  before  only  one  of  them. 

A  coefficient  of  a  letter,  or  of  any  algebraic  quantity,  is 
a  number  placed  immediately  before  it,  or  separated  from 
it  only  by  the  sign  of  multiplication  ;  and  the  coefficient 
shows  how  many  times  the  letter  or  quantity  is  taken. 
Thus,  in  the  expressions,  5  x,  5 .  x,  5  X  2:,  the  coefficient 
of  X  is  5.  When  a  letter  or  quantity  is  written  without 
any  coefficient,  it  is  supposed  to  have  1  for  its  coefficient; 
thus,  X  is  the  same  as  1  x.  Letters,  as  will  be  seen  here- 
after, may  be  used  as  coeflicients. 

The  process  by  which  an  equation  is  formed  from  the 


§>  1.]  EQUATIONS    OF    THE    FIRST    DEGREE.  25 

conditions  of  a  question,  is  called  putting  the  question  intj 
an  equation  ;  and  the  process  by  which  the  value  of  the 
unknown  quantity  is  found  from  the  equation,  is  called 
solving  the  equation, 

3.  A  man  bought  a  certain  number  of  pounds  of  coffee, 
at  10  cents  per  pound,  and  the  same  number  of  pounds 
of  sugar,  at  8  cents  per  pound,  and  the  whole  amounted 
to  ft -02.     Required  the  quantity  of  each. 

Let  X  represent  the   number  of  pounds  of  each  ;   x  lbs. 
of  coffee,  at  10  cents  per   lb.,  will   cost  10  x  cents,  and 
X  lbs.  of  sugar,  at  8  cents  per  pound,  will  cost  8  x  cents. 
Hence,  10  x  -f"  8  x  z=  162.     Reducing  the  first  member, 
18xz=162,  and 

a;i=  9  lbs.  of  each,  Ans. 

4.  A  and  B  hired  a  house  for  $300,  of  which  A  wab 
to  pay  twice  as  much  as  B.    How  much  was  each  to  pay  1 

5.  A  man  is  twice  as  old  as  his  son,  and  his  son  twice 
as  old  as  his  daughter.  Required  the  age  of  each,  their 
united  ages  being  77  years. 

6.  A  farmer  sold  equal  quantities  of  wheat  and  rye,  and 
received  for  both  150  shillings.  Required  the  number  of 
bushels  of  each,  the  wheat  being  sold  at  8  s.  and  the  rye 
at  7  s.  per  bushel. 

7.  A  drover  sold  6  oxen  and  4  cows  for  $320,  receiv- 
ing twice  as  much  for  an  ox  as  for  a  cow.  Required  the 
price  of  a  cow,  also  that  of  an  ox. 

8.  In  a  company  of  160  individuals,  there  were  3  times 
as  many  ladies  as  gentlemen,  and  4  times  as  many  children 
as  there  were  ladies.     Required  the  number  of  each. 

9.  Two  travellers  set  out,  at  the  same  time,  from  two 
towns,  120  miles  apart,  and  travel  towards  each  other  till 
they  meet.  How  long  would  they  be  upon  the  road,  if  one 
goes  5  and  the  other  7  miles  per  hour? 

10.  A  market-man  sold   10  apples,  12  peaches,  and  7 


26  EQUATIONS    0¥     THE    FIRST    DEGREE.  [§  I, 

melons,  for  $2*64,  selling  a  peach  at  twice  the  price  of  an 
apple,  and  a  melon  at  7  times  the  price  of  a  peach.  At 
what  price  did  he  sell  one  of  each  kind? 

11.  A  man  bought  7  bushels  of  potatoes,  5  of  corn,  and 
9  of  wheat,  for  £7  19s.,  giving  twice  as  much  a  bushel 
for  corn  as  for  potatoes,  and  twice  as  much  a  bushel  for 
wheat  as  for  corn.  How  many  shillings  did  he  give  a 
bushel  for  each? 

12.  A  and  B  traded  in  company,  and  gained  $240.  B 
put  in  twice  as  much  stock  as  A.  What  is  each  man's 
share  of  the  gain  ? 

Remark.  It  is  evident  that,  since  B  furnished  twice  as 
much  stock  as  A,  he  should  have  twice  as  much  of  the 
gain. 

13.  A  gentleman  distributed  20  shillings  among  8  beg- 
gars. To  the  first  he  gave  twice  as  much  as  to  the  second, 
and  to  the  second  3  times  as  much  as  to  the  third.  How 
many  shillings  did  he  give  to  each? 

14.  A  man  bequeathed  an  estate  of  $16,000  to  his  two 
sons  and  three  daughters,  directing  that  the  daughters 
should  all  share  alike,  that  the  younger  son  should  have 
twice  as  much  as  one  daughter,  and  that  the  elder  son 
should  have  as  much  as  all  the  daughters.  Required  the 
share  of  each. 

15.  Three  men,  A,  B,  and  C,  built  670  rods  of  fence. 
A  built  7,  B  5,  and  C  4  rods  a  day.  A  wrought  3  times 
as  many  days  as  B,  and  B  wrought  5  times  as  many  days 
as  C.     Required  the  number  of  days  each  wrought. 

16.  A  draper  bought  16  pieces  of  cloth;  3  were  white, 
4  black,  and  9  blue.  A  piece  of  black  cost  twice,  and  a 
piece  of  blue  3  times,  as  much  as  a  piece  of  white.  Re- 
quired the  price  of  a  piece  of  each,  the  cost  of  the  whole 
being  $152. 


X 


^  2,]  EQUATIONS    OF    THE    FIRST    DEGREE.  87 


SECTION  II. 

EQUATIONS  OF  THE  FIRST  DEGREE,  HAVING  UNKNOWN 
TERMS  IN  ONE  MEMBER  ONLY,  AND  KNOWN  TERMS  IN 
BOTH. 

Art.  16.     1.    The  sum  of  the  ages  of  A  and  B  is  50 

years,  and  B  is  10  years  older  than  A.     Required  the  age 
of  each. 

Let  X  represent  A's  age  in  years  ;  then  x  -f-  10  will  rep- 
resent B's  age.     Hence, 

X -{- X -\- 10  i=z  50.     Reducing  the  first  member, 
2x4-10=150. 
Since  the  two  members  of  the  equation  are  equal,  we 
may  now  subtract  10  from  each  member,  and  the  remain- 
ders will  be  equal,  (Ax.  2.)     First  representing  this  sub- 
traction, we  have 

2  X  -f  10  —  10  ==  50  —  10.    Now  reducing,  that  is, 
performing  the  subtraction  represented, 
2  x  =  40  ;  from  which 
a;  z=  20  years,  A's  age  ;  \ 

and  z  +  10  z:^  20  4-  10  =  30  years,  B's  age  ;  ]  ^^^^ 
2.    A  farmer  had  6  more  than  twice  as  many  sheep  as 
cows,  and  the  number  of  his  sheep  and  cows  together  was 
66.     Required  the  number  of  each. 

Let  X  represent  the  number  of  cows ; 
then  2  X  -[-  6  must  be  the  number  of  sheep.     Hence, 
X -f- 2  X -f- 6  z=  66.     Reduce  the  first  member, 
^  3  X  -f-  6  =  66  ;  subtract  6  from  each  member, 

3x-f-6  —  6  =  66  —  6;  reduce  both  members, 
3  X  z=  60,  and 

X  =  20  cows, 

Ans. 


2  X  -j-  6  =  46  sheep 


.} 


28  EQUATIONS    OF    THE    FIRST    DEGREE.  [<§>  2. 

Remark.  In  the  two  preceding  questions,  we  see  that, 
after  reducing  the  first  equation,  we  subtracted  from  both 
members  the  known  term,  which  was  in  the  same  member 
with  the  unknown  quantity.  Let  the  learner  solve  the 
following  problems  in  a  similar  manner,  first  representing 
the  subtraction,  and  afterwards  reducing. 

3.  A  and  B  hired  a  pasture  for  $75,  of  which  A  paid 
$15  more  than  B.     How  much  did  each  pay? 

4.  A  man  and  his  son  could  earn  together,  in  one  day, 
$2'50,  but  the  man  earned  5  s.  more  than  his  son.  How 
much  could  each  earn  1 

5.  A  laborer  wrought  10  days,  having  the  assistance  of 
his  boy  4  days,  and  received  for  the  wages  of  both  $12 
but  the  man  earned,  in  a  day,  $0*50  more  than  the  boy. 
Required  the  daily  wages  of  each. 

6.  A  is  2  years  older  than  B,  and  B  is  3  years  older 
than  C.  Required  the  age  of  each,  the  sum  of  their  ages 
being  68  years. 

7.  A  grocer  gave  $53  for  5  barrels  of  flour  and  4  bar- 
rels of  rice,  paying  $2  a  barrel  more  for  the  rice  than  for 
the  flour.     Required  the  price  of  a  barrel  of  each. 

8.  Divide  $83  between  A,  B,  and  C,  so  that  B  shall 
have  $7  more  than  A,  and  C  $9  more  than  B. 

9.  In  a  certain  town  there  are  10  more  Irish  than 
English,  and  30  more  French  than  Irish.  Required  the 
number  of  each  class,  there  being  710  persons  in  all. 

ii -'10.  A  man  paid  £^  4  s.  for  4  bushels  of  corn  and  5 
bushels  of  wheat,  giving  2  s.  more  a  bushel  for  wheat  than 
for  corn.     Required  the  price  of  a  bushel  of  each. 

If,  in  solving  the  equations  arising  from  the  preceding 
questions,  the  learner  had,  after  representing  the  subtrac- 
tion in  both  members,  reduced  only  that  member  in  which 
the  unknown  quantity  was  found,  he  would  have  perceived 
that  the  resulting  equation  might  have  been  obtained  b)- 
carrying  the  known  term,  which  was  on  the  snrae  side  as 


^  '2.J  EQ,UATIONS    OF    THE    FIRST    DEGREE.  29 

the  unknown  quantity,  to  the  other  member,  and  changing 
its  sign  from  -[-to  — .  Thus,  in  the  second  question,  we 
had  the  equation  3x4-6  =  66.  If  we  represent  the  sub- 
traction of  6  from  each  member,  the  equation  becomes 
3a:-(-6  —  6  =  66  —  6.  Reducing  the  first  member  of 
this,  we  have  3x  =  66  —  6,  which  might  have  been  ob- 
tained from  3  X -]- 6  =:  66,  merely  by  transferring  6  from 
the  first  member  to  the  second,  and  changing  its  sign  from 
+  to  — . 

In  like  manner,  if  we  had  the  equation  Sx:=:x-\-20, 
by  representing  the  subtraction  of  x  from  each  member, 
and  then  reducing  the  second  member  of  the  result,  we 
should  have  3  2:  —  x=z  20,  which  might  have  been  obtained 
from  3x=zx-\-  20,  by  transferring  x,  supposed  to  have 
the  sign  -f"?  from  the  second  member  .  to  the  first,  and 
changing  its  sign  from  -f"  to  — . 

Art.  17.  Removing  a  term  from  one  member  of  an 
equation  to  the  other,  is  called  transposing  that  term,  oi 
transposition.  Hence,  in  an  equation,  any  term  affected  hy 
the  sign  -j-  may  be  transposed,  if  this  sign  be  changed  to 
— ;  for  this  is  subtracting  the  same  quantity  from  each 
member.     (Ax.  2.) 

Art.  18.     1.    Two  men,  A   and  B,  hired  a  farm  for 

$450,  of  which  A  paid  $50  more  than  B.     Required  the 

rent  paid  by  each. 

Let  X  be  the  number  of  dollars  A  paid ; 

then    X  —  50  will  be  the  number  B  paid.     We  have,  then^ 

x-^x  —  50  =  450.     Reducing  the  first  member, 

2  2:  —  50  =  450  ;   adding  50  to  each  member, 

2  x  —  50  +  50  =  450  +  50  ;  reducing,  (Ax.  5,) 

2x  =  500;  and 

x  =  $250,  what  A  paid  ;  , 

Ans 


2:  — 50=  $200,  what  B  paid 


;} 


30  E<^UATIONS    OF    THE    FIRST    DEGREE.  [<§>  2, 

2.  B  is  10  years  older  than  C,  and  A  is  15  years  older 
than  B.  Required  the  age  of  each,  the  sum  of  their  ages 
being  95  years. 

Let  X  represent  A's  age  m  years  ; 
then     X  —  15   will   represent  B's,   and  x  —  15  —  10  will 
represent  C's. 
Hence,  2:  +  x  —  15  +  a:  —  15  —  10  ziz  95. 
Reducing  the  first  member, 
3  a:  —  40  =  95  ;  adding  40  to  each  member, 
3  a;  —  40  +  40  1=  95  +  40  ;  reducing,  (Ax.  5,) 
3  a:  =  135  ;  and 
a;  =:  45  years,  A's  age  ;  ^ 
X  —  15  =1  30  years,  B's  age  ;  >  Ans. 
r  —  15  —  10  =  20  years,  C's  age  ;  ^ 
Remark  1st.     It  will  be  perceived,  in  the  reduction  of 
the  first  equation  of  this  question,  that  subtracting  several 
quantities   separately,  is  equivalent  to  subtracting   their 
sum  at  once. 

Remark  2d.  Most  of  the  preceding  questions  in  this 
section  may  be  performed  in  the  same  way  as  the  last  two 
have  been.  Let  the  learner  solve  the  following  questions 
in  a  similar  manner. 

3.  A  man  performed  a  journey  of  70  miles  in  3  days, 
travelling  4  miles  less  the  2d  day  than  he  did  the  1st,  and 
3  miles  less  the  3d  day  than  he  did  the  2d.  Required  the 
number  of  miles  he  went  each  day. 

4.  A  had  3  times  as  much  money  as  B.  After  A  had 
lost  $20  and  B  $15,  it  was  found  that  they  had,  together, 
$85.     How  much  money  had  each  at  first  ? 

5.  When  oats  were  3  s.  a  bushel  cheaper  than  corn,  a 
stage  owner  paid  $12*50  for  7  bushels  of  oats  and  9  bush- 
els of  corn.     Required  the  price  of  each  per  bushel. 

6.  A  trader  exchanged  20  bushels  of  rye  and  13  bush- 
els of  corn  for  12  bushels  of  wheat,  estimated  at  8  s.  per 


tt 


<§>  2.]  EQ^UATIONS    OF    THE    FIRST    DEGREE.  31 

bushel,  and  .£5  9  s.  in  money.  Required  the  estimated 
price  of  the  rye  and  corn  per  bushel,  corn  being  worth 
2  s.  less  per  bushel  than  rye. 

7.  The  difference  of  2  numbers  is  5 ;  and  7  times  the 
greater,  together  with  10  times  the  less,  makes  171.  Re- 
quired the  numbers. 

8.  There  is  a  pole  consisting  of  two  parts,  the  upper 
being  6  feet  less  than  3  times  the  length  of  the  lower. 
Required  the  length  of  each  part,  the  whole  pole  being  30 
feet  long. 

9.  A  and  B  have  equal  sums  of  money.  A  doubles  his, 
and  B  loses  <£50,  after  which  they  both  together  have 
.£250.     How  much  had  each  at  first  ? 

If,  in  solving  the  equations  arising  from  the  preceding 
questions  in  this  article,  the  learner  had,  after  representing 
the  addition  to  both  members,  reduced  only  that  member 
in  which  the  unknown  quantity  was  found,  he  would  have 
perceived,  that  the  resulting  equation  might  have  been  ob- 
tained by  carrying  the  known  term,  which  was  on  the  same 
side  as  the  unknown  quantity,  to  the  other  member,  and 
changing  its  sign  from  —  to  -f-. 

Thus,  in  the  first  question,  we  had  2  x  —  50  :=:  450. 
If  we  represent  the  addition  of  50  to  both  members,  the 
equation  becomes  2  x  —  50  -}-  59  :=  450  -j-  50.  Reducing 
the  first  member  of  this,  we  have  2  x  ^n  450  -|-  50,  which 
might  have  been  obtained  from  2  x  —  50  =  450,  merely 
by  transferring  —  50  from  the  first  member  to  the  second, 
and  changing  its  sign  from  —  to  -|— 

In  like  manner,  if  we  had  the  equation  7  x  i=  81  —  2  x, 
by  representing  the  addition  of  2  x  to  each  member,  and 
then  reducing  the  second  member  of  the  result,  we  should 
have  7x-|-2xi=81,  which  might  have  been  obtained 
tiom  7  X  zz:  81  —  2  X,  by  transferring  —  2  x  from  the  second 
member  to  the  first,  and  changing  its  sign  from  —  to  4- 


3^  EC^UATIONS     OF    THK    FIRST    DEGREE.  [§2 

Art.  lO,  Hence,  any  term  affected  hy  the  sign  — , 
may  he  transposed  from  one  member  of  an  equation  to  the 
other,  if  this  sign  be  changed  to  -\-  ;  for  this  is  adding  the 
same  quantity  to  each  member. 

Art.  20.  Combining  the  preceding  principle  with  that 
given  in  Art.  17,  we  have  the  following 

GENERAL    RULE    FOR    TRANSPOSITION. 

Any  term  may  he  transposed  from  one  memher  of  an 
equation  to  the  other,  provided  its  sign  be  changed  from 
-(-  to  — ,  or  from  —  to  -f-. 

Art.  21.  1.  B  was  10  years  older  than  A,  C  was  3 
times  as  old  as  A  wanting  20  years,  D  was  30  years  older 
than  B,  and  the  sum  of  their  ages  was  150  years.  Re- 
quired the  age  of  each. 

Let  X  be  A's  age  in  years ;  then  B's,  C's,  and  D's  will 
be,  respectively,  x-\-lO,Sx  —  20,  and  a:  4-10  + 30,  or 
X  -|-  40.     Hence, 

x^x-{-10'i-Sx  —  20-\-x-{-4:0  —  150.     Reduce  the 
first  member, 
6  a:  +  30  =  150  ;  transpose  30, 

6  X  =  150  —  30  ;  reduce  the  second  member, 
6x=::120;   .-. 
X  zz:  20  years,  A's  age  : 
X  +  10  =  30  years,  B's  age  ;  v^  j^^^ 
Sx  —  20  =  40  years,  C's  age  : 
x  -|-  40  =  60  years,  D's  age  ; 
2.    Four  men  commenced  trade  with  equal  stocks.    The 
first  doubled  his  stock  and  =£10  more ;  the  second  doubled 
his  and  £20  pounds  more ;    the  third   tripled  his  stock 
wanting  .£100;  and  the  fourth  quadrupled  his,  wanting 
£250 ;  after  which  they  all,  together,  had  £780.     What 
was  the  stock  of  each,  at  first  ? 


<§>  2.]  EQUATIONS    OF    THE    FIRST    DEGREE.  33 

Let  X  represent  the  number  of  pounds  each  had  at  first. 
Then,  2  x  +  10,  2  2:  +  20,  3  a:  —  100,  and  4  2:  —  250,  will 
represent  their  respective  stocks  after  they  had  made  their 
gains.     Hence, 

2  2:  +  10  +  2x  +  20  +  3a:— 100 +  4x  — 250  =  780. 
Reduce  the  first  member, 

II  a;  ~  320  =  780 ;  transpose  —  320, 
11  x=:  780 +  320;  reduce, 
,    11  2:  =1100;  .-. 

a:  =  ^100,  each  man's  stock  at  first,  Ans, 

3.  A  man  walked  91  miles  in  3  days.  He  went  12 
miles  more  the  2d  day  than  he  did  the  1st,  and  20  miles 
less  the  3d  day  than  he  did  the  2d.  Required  the  extent 
of  each  day's  journey. 

4.  In  a  certain  manufactory  there  are  10  more  than  3 
times  as  many  boys  as  men,  and  40  less  than  4  times  as 
many  girls  as  boys.  Required  the  number  of  each,  there 
being  138  in  the  whole. 

5.  A  farmer  has  a  certain  number  of  plum-trees,  10 
less  than  twice  as  many  peach-trees,  20  more  pear-trees 
than  peach-trees,  and  twice  as  many  apple-trees  as  of  all 
others.  Required  the  number  of  each  kind,  the  whole 
number  of  trees  being  300. 

6.  In  a  school  of  155  scholars,  a  certain  number  learn 
geometry,  twice  as  many  learn  algebra,  and  20  less  than 
twice  as  many  learn  arithmetic  as  algebra.  These  three 
classes  constituting  the  whole  school,  it  is  required  to  find 
the  number  in  each  class. 

7.  B  has  $200  more  than  A,  and  C  has  $300  less  than 
twice  as  much  as  B,  and  they  have,  together,  $1900. 
Required  each  man's  money. 

8.  .  Four  towns  are  situated  in  a  straight  line,  and  in  the 
order  of  the  letters  A,  B,  C,  and  D.  The  distance  from 
B  to  C  is  10  miles  more  than  that  from  A  to  B  ;  and  the 

\\ 


34  EQUATIONS    OF    THE    FIRST    DEGREE.  [<§>  3. 

distance  from  C  'o  D  is  20  miles  less  than  twice  that  from 
B  to  C.  Requir  ^d  the  distances  from  A  to  B,  from  B  to 
C,  and  from  C  U  D,  the  entire  distance  from  A  to  D  being 
210  miles. 

9.  A  drove*  jought  a  certain  number  of  oxen,  at  $30 
each ;  5  more  c  iws  than  oxen,  at  $20  each ;  and  50  less 
than  5  times  >*  many  sheep  as  cows,  at  $5  each.  The 
whole  cost  hiio  $725.     Required  the  number  of  each. 

10.  A  dra^^er  bought  9  yards  of  broadcloth,  12  of  cassi- 
mere,  and  J5  of  silk.  A  yard  of  broadcloth  cost  $5  more, 
and  a  yard  of  silk  $1  less,  than  a  yard  of  cassimere,  and 
the  whole  cost  him  $102.  Required  the  price  of  each 
per  jfard. 


SECTION   III. 

EQUATIONS  HATING  BOTH  KNOWN  AND  UNKNOWN  TERMS 
IN  EACH  MEMBER.  ^ 

Art.  33.  1.  A  certain  man's  age  is  such,  that  if  he 
were  10  years  older,  he  would  be  twice  as  old  as  he  would 
be  if  he  were  10  years  younger.     Required  his  age. 

Let  X  be  his  age  in  years ;  then  x  -|-  10  must  be  double 
x  — 10.     Hence,  2  a:  —  20  —  x  +  10. 

In  solvmg  an  equation  having  one  unknown   quantity, 
we  wish  to  get  all  the  unknown  terms  into  one  member 
by  themselves,  and  it  is  generally  best  to  get  the  unknown 
into  the  first  member,  and  the  known  into  the  second. 
Iij  the  equation  given  above,  transpose  the  —  20, 
2  X  =  z  -)-  10  -f-  20  ;  transpose  x, 
2x  —  xzzi  10  -}-  20 ;  reduce, 
X  =z  30  years,  Ans. 


^3.]  EQUATIONS    OF    THE    FIRST    DEGREE.  35 

Let  us  solve  the  same  equation  written  with  its  mem- 
bers reversed.  < 

X  +  10  :=  2  a:  —  20.     Transpose  10, 

a:  =  2  a:  —  20  —  10;  transpose  2  a;, 
x  — 2a;  =  — 20  — 10;  reduce, 

—  x=L  —  30  ;  transpose  both  members, 
30  =  X,  which  is  the  same  as 
a:i=:30. 
This  last  equation,  x  ==  30,  might  have  been  obtained 
from  —  a:  =  —  30,  merely  by  changing  the  signs  of  both 
members. 

Or  we  might  have  changed  the  signs  of  all  the  terms  in 
the  equation  x  —  2xz=  —  20  —  10,  which  would  then  have 
become  — a:  +  2a:  — 20  +  10,  or  2a:  — a;  =  20+ 10,  the 
same  as  in  the  first  solution  after  transposition.     Hence, 

In  any  equation,  the  signs  of  all  the  terms  may  he 
changed;  for  this  is  equivalent  to  transposing  all  the 
terms. 

This  change  of  signs  should  always  be  made,  whenever 
the  first  member  becomes  minus.  The  student  must 
recollect,  however,  to  change  the  signs  of  all  the  terms, 
otherwise  great  errors  will  ensue. 

2.  A  has  twice  as  much  money  as  B ;  but  if  B  had  «£20 
more,  and  A  .£10  less,  B  would  have  3  times  as  much  as 
A.     Required  the  money  of  each. 

Suppose    B    had   x   pounds ;    then   A  would    have   2  x 
pounds.     If  B  had  <£20  more,  he  would  have  a:  -f-  20  ;  and 
if  A  had  <£10  less,  he  would  have  2  a;  — 10.     Hence, 
x  +  20  =  6  a:  —  30.     Transpose  20, 

X  =  6  x  —  30  —  20  ;  transpose  6  x, 
X  — 6x  =  — 30  — 20;   reduce, 
—  5  X  =  —  50  ;  change  the  signs, 
5x=:50;   .-, 

xiz=<£lO,  B's  money;  _ 

•  An$, 


2x1=  .£20,  A's  money 


;} 


36  EQUATIONS    OF    THE    FIRST    DEGREE.  [<^  3. 

If  the  first  equation  had  been  reversed,  both  members 
would,  after  transposition  and  reduction,  have  been  affect- 
ed by  the  sign  -|-,  without  changing  the  signs. 

3.  A  man  bought  3  horses.  For  the  first  he  gave  $30 
less,  and  for  the  third  $20  more,  than  for  the  second  ; 
moreover,  the  price  of  the  third  was  double  that  of  the 
first.     Required  the  price  of  each. 

4.  Two  numbers  differ  by  20  ;  and  10  more  than  4  times 
the  less  is  equal  to  30  added  to  twice  the  greater.  Re- 
quired the  numbers. 

5.  A  man  has  4  sons,  each  of  whom  is  3* years  older 
than  his  next  younger  brother  ;  and  the  sum  of  the  ages 
of  the  younger  two  is  equal  to  the  age  of  the  eldest. 
Required  their  ages. 

6.  A  and  B  have  equal  sums  of  money ;  and  if  A  gives 
B  $50,  B  will  have  3  times  as  much  as  A  has  lefl.  How 
much  money  has  each  ? 

7.  A  carpenter,  wanting  a  stick  of  timber  of  a  certain 
length,  had  two  sticks  on  hand,  one  of  which  was  6  feet  too 
short,  and  the  other  14  feet  too  long.  Now,  3  times  the 
length  of  the  shorter  was  equal  to  twice  that  of  the  longer. 
What  length  of  timber  did  he  require  ? 

*  8.  The  ages  of  two  boys  differ  by  4  years,  and  3  times 
the  age  of  the  elder  is  the  same  as  5  times  that  of  the 
younger.     Required  their  ages. 

9.  A  gentleman  bought  a  horse  and  chaise.  For  the 
chaise  he  gave  $50  more  than  for  the  horse ;  and  5  times 
the  cost  of  the  chaise  was  the  same  as  6  times  that  of  the 
hrrse.     Required  the  price  of  each. 

10.  Two  towns,  60  miles  from  each  other,  are  in  the 
same  direction  from  Boston,  on  the  same  straight  road, 
and  one  is  5  times  as  far  from  Boston  as  the  other.  Re- 
quired the  distance  of  each  from  Boston.  y; 

11.  A  grocer  has  two  kinds  of  tea,  one  of  which  is 


^  3.]  EQUATIONS    OF    THE    FIRST    DEGREE.  37 

worth  3  s.  per  pound  more  than  the  other  ;  moreover,  4 
pounds  of  the  one  are  worth  as  much  as  7  of  the  other. 
Required  the  worth  of  a  pound  of  each. 

12.  A  boy,  distributing  some  peaches  among  his  com- 
panions, found  that  he  wanted  3  in  order  to  give  them  7 
apiece  ;  he  therefore  gave  them  6  apiece,  and  had  1  peach 
left  What  was  the  number  of  his  companions,  and  how 
many  peaches  had  he  ? 

13.  A  boy  was  sent  to  market  for  a  certain  number  of 
pounds  of  meat.  If  he  bought  beef  at  10  cents  a  pound, 
he  would  have  $0*80  left ;  but  if  he  bought  mutton  at  8 
cents  a  pound,  he  would  have  $1*04  left.  How  many 
pounds  of  meat  was  he  to  purchase,  and  how  much  money 
had  he  with  him  ? 

14.  A  farmer,  having  a  certain  number  of  young  fruit 
trees,  wished  to  set  them  in  rows,  with  a  certain  number 
in  a  row ;  but  he  lacked  6  trees  in  order  to  make  7  rows, 
and  if  he  set  out  6  rows,  he  would  have  2  trees  remaining. 
How  many  trees  did  he  wish  to  place  in  each  row  ?  and 
how  many  trees  had  he? 

15.  B  has  $10  more  than  A,  and  $5  less  than  C ;  more- 
over, 3  times  what  A  and  B  together  have,  is  equal  to 
twice  what  B  and  C  together  have.  How  much  money 
has  each  ? 

16.  A  farmer  sold  his  barley  for  1  s.  less,  and  his  rye 
for  2  s.  more,  per  bushel,  than  he  did  his  corn ;  and  he 
found  that  2  bushels  of  barley  and  5  bushels  of  corn  came 
to  as  much  as  5  bushels  of  rye.  Required  the  price  of 
each,  per  bushel. 

17.  Separate  50  into  two  parts,  so  that  3  times  the  less 
shall  be  equal  to  twice  the  greater. 

Let  X  represent  the  less;  then  50  —  x  will  be  the 
greater. 

18.  A  had  $60,  and  B  $20.    B  borrowed  a  certain  sum 


38  EQUATIONS    OF    THE    FIRST    DEGREE.  [<§>  4. 

of  A,  after  which  he  had  in  all  twice  as  much  as  A  had 
left.     How  much  did  he  borrow  of  A  ? 

19.  When  corn  was  worth  3  s.  a  bushel  more  than  oats, 
a  man  gave  9  bushels  of  oats  and  6  s.  in  money  for  6 
bushels  of  corn.  What  were  the  estimated  prices  of  the 
oats  and  corn  per  bushel  ? 

20.  Two  gentlemen,  each  having  $20  in  his  pocket, 
contributed  to  a  public  charity,  one  giving  twice  as  much 
as  the  other ;  and  it  was  found  that  one  had  remaining  3 
times  as  much  money  as  the  other.  How  much  did  each 
contribute  ? 


SECTION  IV. 


EQUATIONS    CONTAINING    FRACTIONAL    PARTS    OF    SINGLE 
TERMS. 

Art.  33.     1.    Five  sixths  of  a  ton  of  hay  cost  $15. 

What  was  the  price  of  a  ton  1 

Let  X  represent  the  price  of  a  ton,  in  dollars  ;  then  f  of 
a  ton  will,  cost  f  of  x.     Five  sixths  of  x  is  represented 

thus,  f  x,  or  thus,  — ;  the  latter  form  is  the  most  usual. 

6 
This  may  be  read,  either  "  five  sixths  of  x,"   "  5  x  sixths," 

**  a  sixth  of  5  x,"  or  ^^  5x  divided  by  6."     Hence, 
-  -  =  15. 

6 
Now,  we  may  divide  both  members  by  5,  (Ax.  4 ;)   and 
as  a  fraction  is  divided  by  dividing  the  numerator,  we  have 

— =.S,     Since  |^  of  x  is  equal  to  3,  the  whole  of  z  will 

6 

be  6  times  as  much ;  that  is, 

z=:$l8,  Ans. 


^  4.]  EQUATIONS    OF    THE    FIRST    DEGREE.  39 

To  solve  the  equation  another  way,  let  us  resume 

—  =  15.     We  may  first  multiply  both  members  by  6 
6 

and  as  a  fraction  is  multiplied  by  dividing  its  denominator 

we  have 

^— 1=90;  thatis,  5xz=90;  .-. 

x=  $18,  Ans, 
The  latter  mode  of  solution  is  generally  to  be  preferred. 

2.  In  a  certain  village,  ^  of  the  people  are  English,  J 
Irish,  ^  French,  and  the  remainder,  50  in  number,  ar« 
Germans.     What  is  the  whole  population  of  the  village  ? 

Let  X  be  the  number  of  inhabitants.     Then, 

:,=z^+^+|-  +  50.     Multiply  by  2, 

2x  =  x  +  —  +  — +100;  multiply  this  by  2, 

4x  =  2x-^x  +  —  +200;  multiply  this  by  2, 
8x=^4:X'\-2x-\-x-\-  400 ;  reduce, 
8  x  =  7  X  -f-  400  ;  transpose  7  x, 
Sx — 7a;  =  400;  reduce, 
X  =:  400,  Ans. 
The  process,  in  this  question,  would  have  been  shorter, 
if  we  had  multiplied  the  first  equation  by  8  at  once,  since 
we  should  then  have  obtained 

8x=:  —  +  —  +  —  +  400;  that  is. 

2      '      4      '      8      ^ 

Sx  =  4x-\-2x-{-x-\-  400 ;  but  it  is  best  to 
remove  one  denominator  at  a  time,  until  the  learner  has 
become  quite  familiar  with  the  process. 

3.  The  sum  of  f ,  f ,  |-,  and  |^  of  a  certain  number  is 
150      Required  the  number. 

Let  X  represent  the  number.     Then, 


Xl  X         m        O  X         t        O  X  I         I   X  t  P  f\ 

=  loU. 

3      '      4      '      6      '      8 


^ 


40  EQ.IJATIONS    OF    THE    FIRST    DEGREE.  [^4. 

We  are  now  to  multiply  so  as  to  remove,  successively, 
the  denominators.  But  it  is  sometimes  convenient  to 
represent  the  multiplication  of  certain  numbers  in  an 
equation.      Multiply  the  equation  by  3. 

2a:  +  — +  — +  —  iz:3.150;  multiply  by  4, 

Sx-{-9x-\-10x-{-^  =  S,4:.l50;  multiply  by  2, 

16  X  +  18  X  +20  a;  +  21^^  zn  3 .  4 . 2 .  150  ;  reduce, 
75  X  z=  3 . 4 . 2 .  150. 

iNow,  since  dividing  one  factor  divides  a  product,  in  the 
second  member  we  have  only  to  divide  150  by  75.  Divid- 
ing both  members,  we  have 

x  =  3 . 4 . 2 .  2  :=  48,  Ans, 

4.  Four  fifths  of  a  number,  increased  by  2,  is  the  same 
as  I  of  the  number  diminished  by  4.  Required  the 
number. 

5.  The  cost  of  f  of  an  acre  of  land  being  $35,  what 
will  an  acre  cost  at  the  same  rate  1 

6.  Two  numbers  are  to  each  other  in  the  ratio  of  9  to 
7,  and  their  sum  is  112.     Required  the  numbers. 

Remark.  The  ratio  of  9  to  7  means  that  the  greater  is 
^  of  the  less,  or  that  the  less  is  ^  of  the  greater. 

7.  A  man's  age  is  to  that  of  his  wife  as  3  to  2,  and  his 
age  exceeds  hers  by  10  years.     Required  their  ages. 

\^  8.  A  grocer  bought  some  tea,  at  $0*50  per  pound,  and 
I  as  much  coffee,  at  $0*12  per  pound,  and  paid  for  both 
$21*92.     How  many  pounds  of  each  did  he  buy  ? 

9.  In  a  mixture  of  gold  and  copper,  1  ounce  more  than 
f  of  the  whole  was  gold,  and  y^^  of  the  whole  was  copper. 
Required  the  weight  of  the  whole,  and  that  of  each  in- 
gredient. 

10.  In  a  mixture  of  copper,  tin,  and  zinc,  5  oz.  more 
than  f  of  the  whole  was  copper,  2  oz.  less  than  ^  of  the 
whole  was  tin,  and  1  oz.  less  than  :|-  of  the  whole  was  zinc. 


^4.]  EQUATIONS    OF    THE    FIRST    DEGREE  41 

Required  the   weight  of  the  entire  mixture,  and  that  of 
each  ingredient. 

11.  A  man  had  passed  ^  of  his  life  in  Germany,  12 
years  more  than  ^  of  it  in  England,  and  3  years  less  than 
^  of  it  in  France.     Required  his  age. 

12.  A  trader  bought  a  quantity  of  oats,  at  2  s.  a  bushel, 
and  twice  as  much  corn,  at  5  s.  a  bushel.  He  afterwards 
sold,  at  the  same  prices  he.  gave,  -J  of  his  oats  and  f  of  his 
corn  for  79  s.     How  much  of  each  did  he  buy  ? 

13.  If  I  multiply  a  certain  number  by  5,  divide  the 
product  by  7,  and  diminish  the  quotient  by  3,  I  shall  ob- 
tain ^  the  original  number.     Required  that  number. 

14.  A  man's  age,  at  the  time  of  his  marriage,  was  to 
that  of  his  wife  as  3  to  2 ;  but  after  they  had  been  married 
10  years,  3  times  his  age  was  equal  to  4  times  hers. 
Required  their  ages  at  the  time  of  marriage. 

,^^i^l5.  When  a  barrel  of  apples  was  worth  f  as  much  as  a 
barrel  of  flour,  a  farmer  gave  6  barrels  of  apples  and  $3 
in  money  for  3  barrels  of  flour.  Required  the  estimated 
price  of  each,  per  barrel. 

16.  Two  boys  set  out,  at  the  same  time,  and  from  the 
same  place,  to  run  to  a  certain  goal.  One  could  run  only 
I  as  fast  as  the  other,  and  was  5  rods  short  of  the  goal  at 
the  time  his  companion  had  reached  it.  Required  the 
distance  which  they  proposed  to  run. 

17.  What  number  is  that,  §  of  which,  increased  by  6, 
will  be  the  same  as  i^  of  it  diminished  by  3? 


4&  EQ,UATIONS    OF    THE    FIRST    DEGREE.  [<^  5, 


SECTION   V. 

EQUATIONS   CONTAINING   FRACTIONAL  PARTS  OF  QUANTITIES 
CONSISTING    OF    SEVERAL    TERMS. 

Art.  S4:.  1.  A  is  20  years  older  than  B,  and  B's 
age  is  equal  to  -^^  of  A's.     Required  the  age  of  each. 

Let  X  represent  B's  age  in  years ;  then  x  -(-  20  will  rep- 
resent A's.    One  eleventh  of  A's  is  ^        ,  and  —  is  7  times 

11    '  11 

7  X  -f- 140 

as  much ;  that  is,  — — — .     Hence, 
'11  ' 

-  — — +^      Multiply  by  11, 


11 
1 1  x  =  7  X  -[-  140  ;  transpose  7  x, 

11  X  —  7x^=140;  reduce, 

4xz=140;  .-. 

X  z=  35  years,  B's  age ,  \    . 
X  -|-  20  =:  55  years,  A's  age ;  J 
2.    A  had  twice  as  much  money  as  B.    A  gained  $500, 
and  B  $100 ;  then  f  of  B's  money  was  equal  to  y\  of  A's. 
How  much  had  each  at  first  ? 

Let  X  be  the  number  of  dollars  B  had ;  then  A  would 
have  2x  dollars.  After  gaining,  B  would  have  x-|-100, 
and  A  2  X  -|-  500  ;  then  §  of  x  -f- 100  must  be  equal  to  ^ 
of  2  X +  500.     Hence, 

2x±200^  8^^2000^     Multiply  by  3, 

2  X  +  200  =  ^^±^  ,  multiply  by  5, 

10  X  +  1 000  z=  8  X  +  2000 ;  transpose, 
10x  —  8x  =  2000  —  1000;  reduce, 
2xz=:1000;  .-. 

X  z=  $500,  B's  money ;    \ 
2  X  =  $1000,  A's  money;  )  "^^^^ 


§5":]  EQUATIONS    OF    THE    FIRST    DEGREE.  43 

8.  A  man  paid  for  a  horse  and  chaise  $300,  and  f  of 
the  price  of  the  horse  was  the  same  as  ^  the  price  of  the 
chaise  increased  by  $28.     Required  the  price  of  each. 

Suppose  the  horse  cost  2:  dollars;  then  300  —  x  will  be 
the  price  of  the  chaise.     Hence, 

4x  _  1500-5 x^^g      Multiply  by  5, 

4.x  =  ™Q--^^  +  140  ;  multiply  by  7, 

28  X  =  7500  --  25  a:  +  980  ;  transpose  —  25  x, 
28  X  +  25  X  zz:  7500  +  980  ;  reduce, 
53xzz:8480;  .-. 

X  1=  $160,  price  of  the  horse  ;    ^ 
300  —  a:  =  $140,  price  of  the  chaise  ;  )  ^^^* 

4.  A  is  15  years  older  than  B,  and  ^  of  A's  age  is  the 
same  as  J  of  B^s.     Required  their  ages. 

5.  What  number  is  that  to  which  if  6  be  added,  and 
from  which  if  5  be  subtracted,  f  of  the  sum  shall  exceed 
f  of  the  difference  by  5  ? 

6.  What  number  is  that  from  which  if  10  be  sub- 
tracted, and  to  which  if  11  be  added,  f  of  the  difference 
shall  be  the  same  as  ^  of  the  sum? 

«.  7.  A  drover,  having  a  certain  number  of  oxen,  and 
twice  as  many  sheep,  bought  5  oxen  and  sold  2  sheep, 
after  which  he  found  that  f  of  his  number  of  oxen  was 
equal  to  ^  of  his  number  of  sheep.  How  many  of  each 
had  he  at  first  1 

mm,  S.  A  man  agreed  to  work  a  year  for  50  bushels  of  corn 
and  $75  in  money ;  but,  on  account  of  sickness,  he 
wrought  only  10  months,  and  received  the  50  bushels  of 
corn  and  $54^  in  money.  What  was  the  estimated  price 
of  the  corn  per  bushel  ? 

9.  A  man's  age,  5  years  before  his  marriage,  was  f  of 
his  age  5  years  after  his  marriage.  Required  his  age  at 
the  time  he  was  married. 


44  EQUATIONS    OF    THE    FIRST    DEGREE.  [<§>  5. 

10.  A  and  B  had  equal  sums  of  money.  A  lent  B  $5, 
and  had  left  y\  as  much  as  B  then  had.  How  much 
money  had  each  at  first? 

11.  Separate  68  into  two  parts,  such  that  f  of  the 
e^reater  shall  be  the  same  as  |  of  the,  less. 

«..  12.  The  joint  stock  of  two  partners,  whose  particular 
shares  differed  by  c£300,  was  to  the  greater  as  20  to  13. 
Required  the  stock  of  each  partner. 

^-^  13.  A  and  B  began  trade,  A  with  twice  as  much  capital 
as  B.  A  doubled  his  stock  wanting  <£300,  and  B  doubled 
his  stock  and  .£50  over  ;  after  which  A's  stock  was  to  B's 
as  14  to  11.  Required  the  stock  with  which  each  began. 
14.  If  I  multiply  a  certain  number  by  3,  subtract  10 
from  the  product,  and  take  f  of  the  remainder,  the  result 
will  be  once  and  a  half  of  the  original  number.  Required 
that  number. 

^  15.  The  united  ages  of  a  man  and  his  wife  amount  to 
104  years,  and  f  of  the  man's  age  is  the  same  as  ^  of  his 
wife's.     Required  the  age  of  each. 

^  16.  A  man  had  two  horses,  and  a  chaise  which  was 
worth  $200.  The  first  horse  and  chaise  were  together 
worth  twice  as  much  as  the  second  horse ;  the  second 
horse  and  chaise  together  were  worth  twice  and  a  half  as 
much  as  the  first  horse.  Required  the  value  of  each 
horse. 

17.  B  could  walk  5  miles  more  in  a  day  than  A,  and  f 
of  what  A  could  walk  in  9  days,  was  8  miles  more  than 
B  could  walk  in  4  days  How  many  miles  could  each 
walk  daily  ? 

00^  18.  A  farmer  had  a  certain  number  of  turkeys  and 
twice  as  many  geese.  He  bought  5  turkeys  and  sold  10 
geese,  when  he  found  that  f  of  his  number  of  turkeys 
exceeded  half  his  number  of  geese  by  1  How  many  of 
each  had  he  at  first  ? 


^6.]  EQUATIONS    OF    THE    FIRST    DEGREE.  45 


SECTION  VI. 

EQUATIONS  OF  THE  FIRST  DEGREE,  WHICH  REQUIRE  THE 
SUBTRACTION  OF  QUANTITIES  CONTAINING  NEGATIVE 
TERMS. 

Art.  fi5.  I.  Separate  30  into  two  parts,  such  that 
the  less  subtracted  from  5  times  the  greater,  shall  leave  a 
remainder  of  66. 

If  we  knew  that  the  greater  part  was  16,  the  less  would 
be  30  —  16,  or  14.  We  then  have  to  subtract  30  —  16 
or  14,  from  5  times  16,  or  80.  If  we  subtract  14  from  80, 
the  remainder  is  66,  as  required. 

Now,  without  reducing  30  — 16,  we  will  subtract  it 
from  80.  First  subtract  30  from  80 ;  the  remainder  is 
80  —  30,  or  50.  But  we  have  subtracted  too  much  by  16, 
because  we  were  to  subtract  30  —  16 ;  our  remainder  is, 
therefore,  too  small  by  16;  hence  we  must  add  16  to 
80  —  30,  which  gives^  80  —  30  +  16,  or  66,  for  the  true 
remainder. 

Now,  to  perform  the  question,  let  x  be  the  greater  part ; 
then  30  —  x  will  be  the  less.  We  have  to  subtract  30  —  x 
from  5  X. 

Subtracting  30  from  5  x,  we  have  5x  —  30 ;  but  we 
have  subtracted  too  much  by  x,  because  we  were  to  sub- 
tract 30  —  x;  the  remainder  is,  therefore,  too  small  by  x. 
Hence,  we  must  Jidd  a:  to  5  x  —  30,  which  gives  5x  —  30 
-f-  X,  for  the  true  remainder. 

We  have,  then,  according  to  the  conditions  of  the 
question, 

5x  —  30 -(-2:  =  66.     This  equation  solved  gives 


x  ==  16,  the  greater  part ; 
30  —  xz=z  14,  the  less  part 


>  Ans 


16  EQ,UATIONS    OF    THE    FIRST    DEGREE.  [<§»  6 

2.  A  had  twice  as  much  money  as  B.  A  lost  $20,  and 
B  gained  $36.  Then  B's  money,  subtracted  from  $120, 
would  leave  the  same  remainder  as  A's,  subtracted  from 
$114.     How  mfuch  money  had  each  at  first? 

Let  X  represent  B's  money ;  then  2  x  will  represent  A's. 
After  A  had  lost  $20,  he  would  have  left  2  x  — 20  ;  and  B, 
having  gained  $36,  would  have  x  -\-  36.  Subtracting  B's 
money  from  $120,  the  remainder  is  120  —  x  —  36;  sub- 
tracting A's  from  $114,  the  remainder  is  114 — 2x-|-20. 
Hence, 

120  —  a:  —  36  =  1 14  —  2  X  +  20.     Transpose, 

22;  — xzz:  114  +  20  — 120  +  36;  reduce, 
X  =  $50,  B's  money  ;     | 
2  X  zn  $100,  A's  money  ;  /  '^^^' 

In  the  first  question,  the  subtraction  of  30  —  x  from  5x 
gave  5  X  —  30  +  X ;  that  is,  we  changed  the  sign  of  each 
term  of  30  —  x,  and  then  wrote  it  after  5  x.  In  like 
manner,  in  the  second  question,  the  subtraction  of  x  +  36 
from  114,  gave  114  —  x  —  36,  and  the  subtraction  of 
2x  — 20  from  120,  gave  120  — 2  x +  20. 

Art.  96,  It  follows  from  the  preceding  operations 
and  explanations,  that  any  quantity  is  subtracted  by 
changing  the  signs  of  all  its  terms,  and  writing  it  after 
the  quantity  from  which  it  is  to  be  subtracted. 

1.  A  man,  having  50  sheep,  bought  a  certam  number 
more,  after  which  he  sold  twice  as  many  as  he  had  bought, 
wanting  10,  and  found  that  he  had  35  left.  How  many 
did  he  buy  ?  and  how  many  did  he  sell  ? 

2.  Separate  60  into  two  parts,  such  that  the  less  sub- 
tracted from  twice  the  greater,  shall  leave  the  same  re- 
mainder as  the  greater  subtracted  from  5  times  the  less. 

3.  A,  B,  and  C  had  together  c£120.  B  had  ^140  less 
than  3  times  as  much  as  A ;  and  if  B's  money  were  sub- 


<§>  6.]  EQ,UATIONS    OF    THE    FIRST    DEGREE.  47 

traded  from  A's,  the  difference  would  be  C's.    How  much 
money  had  each  ? 

4.  Separate  75  into  two  parts,  such  that  if  the  greater 
be  subtracted  from  60,  and  the  less  from  40,  the  former 
remainder  shall  be  4  times  the  latter. 

5.  A  is  3  times  as  old  as  B ;  but  if  A  were  15  and  B  5 
years  younger,  the  excess  of  A's  age  above  B's  would  be 
f  of  B's  actual  age.     Required  the  age  of  each. 

6.  A  farmer  had  a  certain  number  of  sheep,  each  of 
which  brought  him  two  lambs.  He  then  sold  all  the  old 
ones  except  10,  and  found  the  number  of  sheep  and  lambs 
remaining  was  2^  times  his  original  number  of  sheep. 
Required  the  number  of  sheep  he  had  at  first. 

7.  A  had  3  times  as  much  money  as  B.  A  lost  $50, 
and  B  gained  $50 ;  then  y%  of  B's  money  subtracted  from 
A's  would  leave  $7  more  than  ^  what  A  had  at  first.  How 
much  had  each  at  first  ? 

Let  X  represent  the  number  of  dollars  B  had ;  then  Sx 
will  represent  what  A  had.  A  having  lost  $50,  and  B 
having  gained  $50,  A  would  have  3  a;  —  50,  and  B,  x  -f-  50. 
Then,  according  to  the  conditions, 

3 X  —  50  —  ^+^  =  ^-^^7.     Multiply  by  10, 

30x  — 500  — 3x— 150=:15x  +  70;  reduce, 

27x  — 650zz:15a;-j-70;  transpose, 
27x— 15a:zz=70  +  650;  reduce, 
12xz=720;  .-. 

X 1=  $60,  B's  money  ;    ^ 
32:=  $180,  A's  money ;  ]  ^^^' 
_  .3x4- 150 

In  the  first  equation, '—  is  represented  as  subtract- 
ed; and  when  we  multiply  by  10,  10  times  that  fraction, 
that  is,  the  numerator,  is  to  be  subtracted.  But  since  each 
of  t  le  terms  3  x  and   150,  has  the  sign  -f-,  the  sign  oi 


48  EQUATIONS    OF    THE    FIRST    DEGREE.  [<§>  6 

each  must  be  changed  to  — ,  which  gives  the  result  as 
exhibited  in  the  second  equation. 

Observe  that  when  the  sign  -f-  or  —  is  placed  before  a 
fraction,  it  affects  the  whole  fraction,  and  does  not  belong 

to  any  one  term  of  the  numerator.     Thus, ^^ —    is 

+  3a;+150 

the  same  as ■ ' . 

10 

8.  A  gentleman  has  two  horses  and  one  chaise.  The 
first  horse  is  worth  $50  more,  and  the  second  $50  less, 
than  the  chaise.  If  f  of  the  value  of  the  first  horse  be 
subtracted  from  that  of  the  chaise,  the  remainder  will  be 
the  same  as  if  ^  of  the  value  of  the  second  horse  be  sub- 
tracted from  twice  that  of  the  chaise.  Required  the  value 
of  each  horse  and  of  the  chaise. 

Let  X  be  the  price  of  the  chaise ;  then  x  -|-  50  will  be 
that  of  the  first  horse,  and  x  —  50  that  of  the  second. 
Hence,  from  the  conditions  of  the  question, 

Sx-\-l50_^  lx  —  350 


3 


Multiply  by  5, 


Sx  —  Sx  —  150=zl0x  —  ^^^^^—-^;  multiply  by  3, 

15  a;  — 9^  —  450  =  30  3:  — 35  x  + 1750;  reduce, 

6x  —  450  =:  —  5x-{-  1750 ;  transpose  and  reduce, 
11  2;:=  2200,  .-. 

x  =  $200,  value  of  the  chaise ;       \ 
a:  -f-  50  =:  $250,  value  of  the  1st  horse ;  >  Ans. 
x  —  50  =  $150,  value  of  the  2d  horse ;  ) 

35  X 1750 

In  the  second  member  of  the  .2d  equation, is 

represented  as  subtracted ;  but  35  x  is  supposed  to  have 
the  sign  -[-,  and  1750  has  the  sign  — ;  hence,  when  we 
remove  the  denominator,  3,  we  subtract  by  changing  these 
Bigns,  which  gives  the  result  in  equation  3d. 

From  the  solution  of  the  two  preceding  questions,  we 


^ 


^6.]  EQUATIONS    OF    THE    FIRST    DEC^REE.  49 

see  that,  when  a  fraction,  having  several  terms  in  the  nu- 
merator, is  represented  as  subtracted,  on  removing  the 
denominator,  we  change  the  signs  of  all  the  terms  of  the 
numerator.  If,  however,  the  fraction  is  preceded  by  the 
sign  +>  no  change  is  to  be  made  in  the  signs  of  the  nu- 
merator, on  removing  the  denominator. 

9.  If  from  a  certain  number  I  subtract  10,  and  then 
subtract  f  of  this  remainder  from  the  original  number,  the 
last  remainder  will  exceed  ^  of  the  original  number  by  6. 

.  Required  the  number. 

10.  A  grocer  had  two  casks  full  of  wine,  one  contain- 
ing twice  as  much  as  the  other.  From  the  smaller  leaked 
out  13,  and  from  the  larger  46,  gallons.  He  then  drew 
from  the  smaller  f  as  many  gallons  as  remained  in  the 
larger,  and  from  the  larger  f  as  many  gallons  as  remained 
in  the  smaller ;  after  which  there  remained  in  the  larger  1 
gallon  more  than  in  the  smaller.  How  many  gallons  did 
each  cask  hold  ? 

11.  A  and  B  are  of  the  same  age;  but  if  A  were  10 
and  B  5  years  younger,  f  of  B's  age  subtracted  from  A's, 
would  leave  the  same  remainder  as  if  |  of  A's  were  sub- 
tracted from  B's.     Required  their  ages. 

12.  A  man,  having  a  lease  for  100  years,  on  being  asked 
how  much  of  it  had  transpired,  said,  that  f  of  the  time 
past  subtracted  from  the  time  to  come,  would  leave  the 
same  remainder,  as  if  -^^  of  the  time  to  come  were  sub- 
tracted from  the  time  past.  How  many  years  of  the  lease 
had  transpired? 

13.  There  is  a  certain  number  such,  that  if  25  be  sub- 
tracted from  it,  and  if  it  be  subtracted  from  125,  f  of  the 
first  remainder  subtracted  from  f  of  the  second,  will  leave 
10  more  than  ^  of  the  original  number.  Required  thai 
number. 

4 


60  MULTIPLICATION    OF    MONOMIALS.  [^  7 


SECTION   VII. 

MULTIPLICATION    OF    MONOMIALS 

Art.  37.  In  pure  algebra,  letters  are  generally  used 
io  represent  known  as  well  as  unknown  quantities.  We 
shall  now  treat  of  operations  upon  quantities  purely  alge- 
braical. 

It  is  to  be  observed,  that  the  addition,  subtraction,  mul- 
tiplication, 6lc.  of  algebraic  quantities  cannot,  strictly 
speaking,  be  performed,  in  the  same  sense  as  they  are  in 
arithmetic,  but  are,  in  general,  only  represented ;  these 
representations,  however,  receive  the  same  names  as  the 
actual  operations  in  arithmetic. 

A  monomial  is  a  quantity  consisting  of  a  single  term;  as, 

a,  hd,  or  -,  (Art.  15.) 
y 
A  binomial  is  a  quantity  consisting  of  two  terms ;  as, 

«  +  m,  or  X  —  y, 

A  trinomial  is  a  quantity  consisting  of  three  terras ;  as, 
a-\-h  —  c. 

Polynomial  is  a  general  name  applied  to  any  quantity 
containing  several  terms. 

Art.  38.  The  product  of  monomials  is  expressed  by 
writing  them  after  each  other,  either  with  or  without  the 
sign  of  multiplication  ;  thus,  a,  b,  a  X  b,  or  ab,  signifies 
that  a  and  b  are  multiplied  together.  The  last  form,  viz. 
a  b,  is  generally  used.  In  like  manner,  the  product  of  a, 
6,  c,  and  d,  is  ab  c  d. 

The  order  of  the  letters  in  a  product  is  unimportant ; 
tKiui  rthis  the  same   rs  b  a.     This  will  be  manifest,  if 


^7.]  MULTIPLICATION    OF    MONOMIALS.  51 

numbers  are  put  instead  of  the  letters.  Suppose  a  rz  4^ 
and  6  zz=  5 ;  then  a  6  =  4 .  5  =  20,  and  6  a  =  5  .  4  zzi  20. 

Hence,  each  of  the  expressions,  abc,  acb^bca^bac, 
cba,  and  cab,  is  the  product  of  a,  b,  and  c.  For  the 
sake  of  uniformity,  however,  the  letters  of  a  product  are 
usually  written  in  alphabetical  order. 

If  Sab  and  5  mn  were  to  be  multiplied  together,  we 
might  write  the  quantities  thus,  Sab  5mn;  but,  s»nce  the 
order  of  the  factors  is  unimportant,  the  numerical  factors 
may  be  placed  next  to  each  other  ;  thus,  S,5abmn, 
Now,  performing  the  multiplication  of  3  by  5,  we  have 
^5abmn. 

But  it  would  be  erroneous  to  write  S5abmn  as  the 
product  of  S  ab  and  Smn,  because  the  value  of  a  figure 
depends  on  its  place  with  regard  to  other  figures.  If  we 
would  represent  the  multiplication  of  the  figures  also,  they 
must  be  separated  either  by  letters  or  by  the  sign  of  mul- 
tiplication ;  SiS,  SabSmn,  S  .5abmn,  or  S  X  5ctb?nn, 

In  like  manner,  the  product  of  2  am,  Sxi/,  and  5cd,  is 
30  a  c  dmxy;  that  of  7,  2  b,  and  4  c?,  is  56  b  d. 

We  infer,  from  the  preceding  examples,  that  the  product 
of  several  monomials  consists  of  the  product  of  the  co« 
efl5cients  and  all  the  letters  of  the  several  quantities. 

1.  Multiply  2cd  by  Sax, 

2.  Multiply  Spq  by  7. 

3.  Multiply  4  «  by  3  6. 

4.  Multiply   17  6'  by  2fnn, 

5.  Multiply  5  ax  by   11  y. 

6.  Multiply  7  p  q  by  4  r  5. 

7.  Multiply  9gh  by  llx. 

8.  Multiply  20  by  Sxt/, 

9.  Multiply  ISmx  by  7  z. 

10.  Multiply  2ef  by  5  x, 

11.  Multiply  25  by  Sabc. 

12.  Multiply  mm  by  mmm. 


5S  MULTIPLICATION    OF    MONOMIALS.  [<5>  7 

In  the  last  example,  the  product  is,  according  to  what 
has  been  shown  above,  mmmmm.  But  when  the  same 
letter  occurs  several  times  as  a  factor  in  any  quantity,  in- 
stead of  writing  that  letter  so  many  times,  it  is  usual  to 
write  it  once  only,  and  place  a  small  figure,  a  little  ele- 
vated, at  the  right,  to  show  how  many  times  that  letter  is 
a  factor.  Thus,  instead  of  mmmmm,  we  write  m^.  In 
the  same  manner,  we  write  a^,  instead  of  aaaaaa. 

In  all  cases,  a  product  contains  all  the  factors  of  both 
multiplicand  and  multiplier.  In  the  12th  example,  m  is 
twice  a  factor  in  the  multiplicand,  and  three  times  in  the 
multiplier  ;  the  product,  therefore,  must  contain  it  five 
times  as  a  factor ;  that  is,  m^  multiplied  by  m^,  gives  m^ 
The  product  of  2«^63  and  ^a^b^  is  6a^6'^,  because 
each  letter  must  be  found  as  many  times  a  factor  in  the 
product  as  it  is  in  both  multiplicand  and  multiplier. 

Art.  99.  The  small  figure,  placed  at  the  right  of  a 
letter,  is  called  the  index  or  exponent  of  that  letter,  and 
affects  that  letter  only,  after  which  it  is  immediately  placed. 
An  exponent,  then,  shows  how  many  times  a  quantity  is 
used  as  a  factor. 

Quantities  written  with  exponents  are  called  powers  of 
those  quantities ;  thus,  m^  is  called  the  second  power  of 
m,  m^  the  third  power,  m^  the  fifth  power,  &lc.  ;  and  when 
a  quantitv  is  written  without  any  exponent,  it  is  supposed 
to  have  1  for  its  exponent,  and  is  called  the  first  power  of 
that  quantity ;  thus,  m,  which  is  the  same  as  m^,  is  called 
the  first  power  of  m. 

Sometimes  the  second  power  is  called  the  square,  and 
tie  third  power  the  cube,  of  a  quantity  names  which, 
though  derived  from  geometry,  are,  for  the  sake  of  con- 
ciseness, very  convenient  in  algebra. 

Thus,  m^  is  read  m  square,  m^  is  read  m  cube. 


«§»  7. J  MULTIPLICATION    OF    MONOMIALS.  58 

Figures  are  also  frequently  written  with  exponents; 
thus,  31  =  3 ; 

32  z=  3  .  3  z=  9  ; 
^  33  =  3 . 3 . 3  =  27  ; 

34z:z3.3.3.3zz:81. 

Exponents  must  be  carefully  distinguished  from  coeffi- 
cients ;  for  4 «  and  a^  are  very  different  in  their  value. 
Suppose  «  to  be  5 ;  then  4  a  =  20,  and  a^  m  5  . 5  . 5  . 5  z= 
625. 

Art.  30.  From  what  precedes,  we  derive  the  fol- 
lowing 

RULE    FOR    THE    MULTIPLICATION    OF    MONOMIALS. 

Multiply  the  coefficients  together^  and  winte  after  their 
product  all  the  different  letters  of  the  several  quantities ^ 
giving  to  each  letter  an  exponent  equal  to  the  sum  of  its 
exponents  in  all  the  quantities, 

1.  Multiply  Sa^m  by  4:a^m^. 

In  this  question,  the  product  of  the  coefficients  is  12 ; 
the  sum  of  the  exponents  of  a  is  5,  and  the  sum  of  the 
exponents  of  m  is  6.     The  answer,  therefore,  is  12  a^  m^. 

2.  Multiply  a^  b^  by  Sa  b^. 

In  this  example,  the  product  of  the  coefficients  is  1 . 3 
or  3 ;  the  sum  of  the  exponents  of  a  is  3,  that  of  the  ex- 
ponents of  6  is  5  :  the  answer,  therefore,  is  3  a^  b^, 

3.  Multiply  7 ab  by  2ac, 


4. 

13  c2  by  4c^d. 

5. 

11  a?m^  by  3a*m. 

6. 

25  X  by  4x5y. 

7. 

2cT  by  c^d. 

8. 

8a^b^c*  by  dab^c. 

9. 

m^  by  Qa^mK 

54 


MULTIPLICATION    OF    MONOMIALS. 


[^7. 


10. 

Multiply  17x2  by  Szy^. 

11. 

(i 

a'^d^  by  adc. 

12. 

(C 

Sab^c 

by  4  a3  62  ^2. 

13. 

(( 

SOpq^ 

by  p^  q. 

14. 

(i 

7m2/i3 

by  am^. 

15. 

'Find  the  product  of  3  a,  4  d,  and  7  a^  d. 

16. 

it 

"  5m2,  3w,  and  6  c  ^3^ 

17. 

it 

"   10  X,  3y2,  and  4xy. 

18. 

it 

"  2,  7  a,  and  a^. 

19. 

it 

"  5,  6^3,  and  ^m^w^. 

20. 

it 

"  p^j  2p,  and  3jo^g'. 

21. 

it 

((         (( 

''  a^  a  62,  and  a3  6c. 

22. 

if 

(         ti 

**  x7/^z^,  x^ij^z^,  and  x^. 

23. 

it 

t         it 

**  2px,  2^2x2,  and  2^3  z^. 

24. 

ti 

*'  36,  4c,  and  36c2e;3. 

25. 

it 

"  z3,  5»2,  and  2xyz, 

26. 

it               4 

"  40m^w,  2  7^3^  and  xyz. 

Art.  31.     What  is  the  value  of  3  w^  62,  if  a  z=  2  and 
6=5? 

Putting  these  numbers  instead  of  the  letters,  we  have 
3a3  62  —  3.23.52=:3.2.2.2.5.5z=  600,  Ans. 

Let  the  learner  find  the  value  of  the  following  algebrai- 
cal expressions,  supposing  a=l,6z=2,  c  =  3,  and  g?=:  5. 

1.  6a26.  5.    7a6J2.  9.   21^2^3. 

2.  4«362.  6.   5a56c2.  10.   562crf. 

3.  9a6c2.  1,   Sashed,  11.    136^6^. 

4.  2c^d.  8.   a4  63f2.  12.    15a2  62c2J2. 


^  8  ]  REDUCTION    OF    SIMILAR    TERMS.  55 


SECTION  VIII. 


REDUCTION    OF    SIMILAR    TERMS. 


Art.  33.  Similar  terms  are  those  which  are  entirely 
alike  with  regard  to  the  letters  and  exponents, 

N.  B.  Similarity  of  terms  is  wholly  independent  of  the 
algebraic  signs  -j-  and  — ,  the  numerical  coefficients,  and 
the  order  of  the  letters.  Thus,  a^  h  and  5  a^  6  are  simi- 
lar ;    so  also  are  ^x^y^  and  5y^x^. 

Any  polynomial  which  contains  similar  terms,  may  al- 
ways be  reduced  to  a  smaller  number  of  terms.  Of  this 
the  student  has  already  had  numerous  examples  in  the 
reduction  of  equations* 

Let  us  take  the  polynomial  2  a^  ^3  _|_  4  ^3  ^^3  _[_  7  ^2  53  — 
m^  n^.  The  two  terms  2  d^  P  and  7  a^  b^  are  similar,  and 
it  is  evident  that  7  times  a  quantity  added  to  twice  the 
same  quantity,  make  9  times  that  quantity  ;  therefore,  in- 
itead  of  2  a^  53  _|.  7  ^2  53^  ^.^  niay  write  9a^b^,  Again, 
Im-^w^  and  — m^n^  are  similar,  and  4  times  a  quantity 
iiminished  by  once  the  same  quantity,  leaves  3  times  that 
juantity  ;  hence,  instead  of  4  m^  n^  —  m^  n^,  we  may  put 
Im^n^.  By  these  reductions  the  polynomial  ^a^b^-^ 
i  m^  n^ -{- 7  a^  b^  —  m^  n^  becomes  9a^b^-\-'Sm^  n?. 

As  a  second  example,  take  the  polynomiar  5  62^2  — 
:imn^-\-  17  62  ^2  —  10  m  n^.  Here,  5  62^2  +  1762  ^2  ^ 
22 62 c2,  and  —2mji^—10mn^  =  —  12 m n^  ;  so  that  the 
given  quantitv,  when  reduced,  becomes  22  b?  c^ —  12  m  w2. 
^For  a  third  example,  take  10,a2  63 -j-3a3  62  — 6a2  63-[- 
3  >  63  —  7  ah^  _  5  a2  63  -|_  6  a3>  —  aU^.  First  unite  the 
terms  of  one  kind,  having  the  sign  +;  10  a'^  b^ -\- S  a^  b^ 
=:.  i3a2  63  .  now  unite  the  terms  of  the  same  kind,  having 


56  REDUCTION    OF     SIMILAR    TERMS.  i'^^' 

the  sign  —  ;  —  6  a^b^  —  5  a^  6^  =:=  —  1 1  a^  53  hence, 
lOa^b^-^'Sa^b^  —  Qa^b^  —  Sa^b^zzzlSa^b^  —  lla^b^ 

—  2a^  b^.    In  like  manner,  Sa^b^-^da^  b^  =  9a^  b^,  and 

—  7  a^  52  —  a^b'^zzz  —  8  a^  6^  ;  hence,  instead  of  these 
four  terms,  we  have  9  a^h^  —  8  a^  52^  which  is  the  same 
as  a^  b^     The  given  quantity,  therefore,  reduced,  becomes 

Whenever  the  total  of  the  negative  terras  exceeds  that 
of  the  similar  positive  terms,  we  take  the  difference  of  the 
two  amounts,  giving  this  difference  the  sign  — . 

Suppose  we  have  to  reduce  6  -j-  3  x^  y3  -j-  7  x^  y3  — 
10  x^  y^  —  5  x^  r/^.  Uniting  the  similar  positive  and  nega- 
tive terms  separately,  we  have  6  -}~  10  ^^^'^  —  15  x^  y'^  ; 
here  we  have  10  times  1^  y"^  added,  and  15  times  the  same 
quantity  subtracted,  which  is  the  same  as  subtracting  5 
times  that  quantity  ;  hence,  instead  of  10  a:^  y^  —  X5  x^  y'^, 
we  may  put  —  5  x^  y3^  ^nd  the  original  quantity  becomes 
b  —  5  x^  2/3.  \ye  g^^  ^^^  ^^jg  jj^g^  \^^xm  might  have  been 
obtained  by  subtracting  10  from  15,  and  putting  before  the 
remainder  the  sign  — ,  and  after  it  the  common  letters 
with  their  proper  exponentsr^ 


Art.  33.  From  the  preceding  examples  we  derive  the 
following 

RULE    FOR    THE     REDUCTION    OF    SIMILAR     TERMS. 

Unite  all  the  similar  terms  of  one  hind  affected  with  the 
sign  -j-,  by  adding  their  coefficients,  and  writing  the  sum 
before  the  common  literal  quantity  ;  unite,  in  like  manner ^ 
the  similar  quantities  of  the  same  kind  affected  with  the 
sign  — ;  then  take  the  difference  between  these  two  sums, 
and  give  to  the  result  the  sign  of  the  greater  quantity. 


^8J  REDUCTION    OF    SIMILAR    TERMS.  57 

Remark,    To  prevent  errors,  it  is  advisable  to  mark  the 
terms  as  they  are  reduced. 

Let  the  learner  reduce  the  follovv^ing  quantities 

1.  «_|-3«-|_5a_}_66  +  6  +  2a.   11^4-14-  . 

2.  lOx  — 9y  +  172:  — 6x  — Sx  +  Sy  +  lOy.^^"^-^/^ 

3.  3aa:  +  4  6a:  — 6«x  +  116  2:  — 2«x  +  36a;./f ^y^--*"^y 

5.  3x2y4-6a:y2_3a:2y_j_5a:y2_|.3^22:.  -f-^jlj 

6.  m2+10a263_|_24«464-2a263_8a2^,3_46ift_ 
4a253_20an.      .yy^."^ 

7.  8a6c  +  10x2y— 4a6c  — 10a:y2-|_3a6c  — 42:2y 
+  12y2x  +  a6c.    K^  #  f,  -^  -  ^  /'  '^  - 

8.  12a2  63c  +  10a63c  — 6a2  63i;^8a63c  — 2a36c 

+  4m  — 20«63c.    ^^lV'c   -tct-1.  -Zaf^t^i^^-r 

6xy2_4x2y_i6xy2_2a6c.  ;:/-^^^  ^J/  f^^/(^  ^5-^ 

10.  x2_j.3  2;3_4  2:2_j_9a;3^6x2  3^a^_lpx2^-_'9a;3_|.     ^ 
5x2y2_|_i5a;3_i7a,2.     /       V  i  '        :  -:.  ^    -^  -  ^ 

11.  3ww  — 6xy2>|_6xz— I3xy2_i0  2:2;— 2x3^2_|. 
4xy2_|_2x;2— lly2a;. 

12.  17j>2^2_^3«5c  — 2i?y— 20p2^2_3«5c  + 
6pg  +  21p2^2_i4p^^l7. 

13.  120  +  w2  w3  _  6  a;  y  _f.  15  ^2  ;i2  _  24  X  y  +  5  — 
m3  w  —  16  +  4m3/t. 

14.  7x2y  +  3  +  4x  — 10x2y  — 17-f  3x— 14x2y  + 
6  +  2x  +  9-|-13x2y  +  xy2. 


58  ADDITION.  \^  9. 


SECTION  IX. 


ADDITION. 


Art.  34.  The  addition  of  positive  monomials,  as 
has  already  been  seen,  consists  in  writing  them  after  each 
other,  putting  the  sign  -f-  before  each  except  that  placed 
first,  which  is  understood  to  be  affected  with  the  sign  -|-. 
Thus,  to  add  a,  6,  c,  and  d,  we  write  a -\- b -\'  c -{- d. 

Art.  3S.  The  addition  of  the  two  polynomials,  a  -|- 
b  and  c  -f-  c?  -f-  ^>  ^^  which  all  the  terms  are  affected  with 
the  sign  -(-,  gives  a-\-b-\-c-\'d-{-e;  for  adding  these 
two  quantities  is  the  same  as  adding  the  separate  terms  of 
which  they  are  composed.  As  an  example  in  numbers,  let 
us  add  10  +  3  and  6  +  9.  Now,  10  +  3  is  13,  and  6  +  9 
is  15,  and  the  sum  of  13  and  15  is  28 ;  also  10  +  3  +  6 
+  9  is  28. 

But  if  some  terms  of  the  quantities  to  be  added  have 
the  sign  — ,  these  terms  retain  that  sign  in  the  sum.  This 
is  easily  seen  in  the  case  of  numbers.  Let  us  add  9  —  3 
to  13  ;  9  —  3  is  6,  and  6  added  to  13  gives  19.  But  if  we 
first  add  9  to  13,  which  is  expressed  thus,  13  +  9,  the  sum 
is  too  great  by  3,  because  we  were  to  add  only  9  —  3,  or 
6  ;  we  must,  therefore,  subtract  3  from  13  +  9,  which 
gives  for  the  true  sum  13  +  9  —  3,  or  19. 

Let  us  now  add  a  —  b  to  m.  Adding  a  to  m,  we  have 
m-\- a;  this  result  is  too  great  by  b,  because  we  were  to 
add  only  a  —  6,  which  is  less  than  ahy  b;  therefore,  after 
having  added  a,  we  must  subtract  6,  which  gives  for  the 
correct  result  m-\-a  —  b. 

We  see,  in  the  foregoing  examples,  that  in  adding  we 


<§>  9.]  ADDITION.  59 

merely  write  the  quantities  after  each  other,  without  any 
change  in  the  signs. 

As  another  example,  ^et  it  be  required  to  add  the  poly- 
nomials a2_|.3a6-f3  62  — c2,   and   oa^  —  7 ab-^lOb^ 

—  6c^.  The  sum  is  a^ -{-Sab -\-Sb^  —  c^ -]- 5  a^ — 
7  ab-\- 10  b^  —  6  c^ ;  but  this  result  contains  similar  terms, 
and  may  be  reduced  ;  it  then  becomes  6a^  —  4  a  6  -f- 13  6^ 

—  7  c^,  which  is  the  sum  in  its  simplest  form. 

Art.  30.  From  what  precedes,  we  have  the  fol- 
lowing 

RULE    FOR    THE    ADDITION    OF    ALGEBRAIC    QUANTITIES 

Write  the  several  quantities  one  after  another,  giving 
to  each  term  its  original  sign,  and  then  reduce  the  similar 
terms. 

Remember  that  those  terms  which  have  no  sign,  ar 
supposed  to  have  the  sign  -|-. 

1.  Add  3  a,  7  6,  6  c,  and  4  a +  2  6. 

2.  Add2a^4m24-3wiw,  and7a2-|-6mw. 

3.  Add  4 a,  3a  —  2c,  and5m  +  7c 

4.  Add  a2_[_2a6  +  62,  and3a2  — 2a6  +  462. 

5.  Add  3x^  — 15x1/ +  ny^,  15 xy,  and  20x2— 13/ 
+  m2. 

6.  Add  4c2  +  5m2,  3^2  — 2  c^,  15  c2— 10^2,  and  x- 
^Sij^  —  c^, 

7.  Add  —20  +  3x2,  17x2  +  5y2,  45  —  7x2,  and  x** 

—  3y2. 

8.  Add— 12a2x  — 20ax2  +  8x3,  and  24a3  +  40ax5» 

—  16a2x. 

9.  Add  a^  +  a^  c2  +  a2  c^,  and  c^  —  a^c^  —  a^  c\ 

10.  Add  x3-f3ax2  +  4a2x-j_2,  and  4x3  — 30x2  + 
10a2a:_l. 


60  SUBTRACTION.  [<§>  10. 

11.  Add8a^b—5ab^  —  8abc  +  4hc^,  and  6ab^-^ 
2a^b  —  abc  —  4.bc^. 

12.  Add  6ab-{-2ac  —  Sbc,  Abd—lab,  and  6bc 
■■\-5b d  —  3  ac, 

13.  Add  10  +  a;2,  30~3y^  4.a^b^'^Sc^,  11x2  + 
7i/^,  and  23  — a2  62. 

14.  Add  17  «2  62  _  10  a  62  c  +  5  a  6  c2,  4£>  a2  62  — 
10  «  6  c2,  and  25  +  10  a  62  c. 

15.  Add  a^  ^  wz  -f-  «  p  wz^^  6  -\-'  S  a  p^m  —  4t  a  p  m^, 
7a3p  m -f- a^  m  ji -f- 6  a  mp2^  and  lOa^m^  —  Samp^ 
—  12  a^  mp. 

16.  Add  a262  — 25  +  x2,  3a2  62  +  y2_4a;2^  75_ 
9a2 62  +  3^2,  and45x2y2^3a;2_9. 


SECTION   X. 


SUBTRACTION. 


Art.  S7»  We  have  already  seen  that  the  subtraction 
of  a  positive  monomial  consists  in  giving  it  the  sign  — , 
and  writing  it  either  after  or  before  the  quantity  from 
which  it  is  to  be  subtracted.  Thus,  to  subtract  6  from 
a,  we  write  a  —  6,  or  — 6-|-«. 

If  we  have  to  subtract  a  polynomial  in  which  all  the 
terms  are  affected  with  the  sign  -f-,  it  is  plain  that  each 
of  the  terms  must  be  subtracted,  that  is,  the  sign  of  each 
term  must  be  changed  to  — . 

As  an  example  in  figures,  if  it  is  required  to  subtract 
7  +  3  from  18,  we  must  subtract  both  7  and  3;  thus,  18 
—  7  —  3,  which  reduced  becomes  8.  This  result  is  cor- 
rect, because  7  +  3  is  10,  and  10  taken  from  18  leaves  8. 


§»  lO.j  SUBTRACTION.  .  61 

In  like  manner,  to  subtract  m-\-7i  from  a,  we  write  a  — 
m  —  n,  changing  the  signs  of  both  7n  and  n. 

If  some  of  the  terms  in  the  quantity  to  be  subtracted 
have  the  sign  — ,  the  signs  of  those  terms  must  be 
changed  to  -|— 

As  an  example  in  numbers,  let  us  subtract  7  —  4  from 
12 ;  7  —  4  is  3,  and  3  subtracted  from  12  leaves  9.  Now, 
lo  subtract  7  —  4  without  reducing  it,  if  we  first  subtract 
7,  which  is  expressed  thus,  12  —  7,  we  subtract  too  much 
by  4,  and  the  remainder,  12  —  7,  or  5,  is  too  small  by  4; 
consequently,  after  having  subtracted  7,  we  must  add  4, 
which  gives  12  —  7  4-4,  or  9,  for  the  true  remainder. 

In  like  manner,  a  —  b  subtracted  from  c,  gives  c  —  a 
-(-  h.  For,  if  a  be  subtracted  from  c,  we  have  c  —  a;  but 
since  a  is  greater  than  a  —  6  by  6,  we  have  subtracted  too 
much  by  b,  and  the  remainder,  c  —  a,  is  too  small  by  6 ;  we 
must  therefore  add  b  to  c  —  a,  and  we  have  c  —  a-j-b  for 
the  true  result. 

For  another  example,  let  us  subtract  a^  —  2ab-\-Sm^ 
from  5a^-^4:ab  —  2m2.  As  in  the  previous  examples, 
it  may  be  shown  that  the  terms  in  the  quantity  to  be  sub- 
tracted must  have  their  signs  changed.  Making  this 
change  in  the  signs,  and  then  writing  the  quantities  after 
each  other,  we  have,  for  the  remainder,  5a^-\-4tab  — 
2 m^  —  a^^2ab  —  Srn^.  This  remainder  is  correct,  but 
it  may  be  reduced,  and  it  then  becomes  4  a^  -["  ^  ^  ^  — 
5  m^,  which  is  the  remainder  in  its  simplest  form. 

Art.  38,  From  what  precedes,  we  derive  the  fol- 
lowing 

RULE    FOR    THE    SUBTRACTION    OF    ALGEBRAIC    QUANTITIES. 

Change  the  signs  of  all  the  terms  in  the  quantity/  to  be 
subtracted,  from  +  to  — ,  or  from  —  to  -^,  and  write  it 


62  SUBTRACTION.  [<^  10. 

after  that  from  which  it  is  to  be  subtracted ;  then  reduce 
similar  terms 

1.  From  na^b  —  6ab^,  subtract  4 a^ 6  —  7 a 6^ -f  m'-^ 
Changing  the  signs  of  all  the  terms  in  the  latter  quan- 
tity, and  writing   the  result   after  the   former,  we   have 
Ua^b  —  6ab^  —  4:a^b-{'7ab^  —  m^,  which,    reduced, 
becomes  7  a^b-]-  ab^  —  m^,  Ans. 

2.  From  a^-{-2ab'^b^,  subtract  a^  —  2ab-\-b^. 

3.  From  3m2-|-9mc  +  2c2,  subtract  Sm^  —  9mc-\' 
2c2. 

4.  From  6 a^b^-\- 17 abc,  subtract  Sa^b^  — 5 ab c. 

5.  From  25  -(-  ^,  subtract  25  —  3  x. 

6.  From  x^-f  27,  subtract  x^  — 27. 

7.  From  10  a^b^c — 13mw-f-4m2,  subtract  — 15 mn 
-j-  6  a2  62  c  _  3  m2. 

8.  From  Sab  —  2cd-\--5ac  —  7 ad,  subtract  3 a 6  + 
4i  c  d -{-' 5  a  c. 

9.  From  Sbd-\-2a-^m,  subtract  26c?  —  3a  —  b. 

10.  From  7 a^b^ c -\-5ab^c^  —  9ab c -{-m,  subtract 
2a^b^c  —  4ab^c^  —  8abc  —  2m. 

Art.  39.  The  subtraction  of  a  polynomial  is  indi- 
cated  by  enclosing  it  within  brackets,  or  a  parenthesis, 
and  placing  the  sign  —  before  the  whole.  Thus,  3  m  — 
[a^  —  7a6-|-^],  or  Sm  —  (a^  —  7 ab-\-m),  indicates  the 
subtraction  of  a^  —  7  a  6  -|-  m  from  3  m.  Performing  this 
subtraction,  recollecting  that  a^  is  affected  with  the  sign 
-(-,  we  have  2  m  —  a^-\-7  ab. 

Let  the  learner  perform  the  subtraction  indicated  in  the 
following  examples, 

1.  3«2^6  — (a2_i06  +  c2). 

2.  4^2  — n2—(7a6  — 6^2  +  3/12). 

3.  a24-2ac4-c2— (a2_2ac  +  c2). 


<§>10.J  SUBTRACTION.  63 

4.  5abc-^Uvi^  —  {10abc  —  Sm^  —  Scd). 

5.  — 7  ax^  i/^-^-Sx^  7/  —  {U  x^  1/  —  10  x^  1/^  —  5  ab  c). 

6.  a^-{-6ax-\-Wax^  —  {a^  —  6ax  —  10ax^). 

fs.    7xz  —  Sxz^-i-Sz^  —  {Sxz^  —  7xz-{-Wz^), 
/  9.    4a:2y2_4a;y3_i_(2x2y2_4a;^3_^5),      * 

1    10    80a6  — 15ac2  +  15m2a;  — (15ac2+30a6  — 
U5  m^  x). 

Art.  40 •  It  is  often  useful  to  reverse  the  p^-ocess  in 
the  last  article,  and  place  part  of  a  polynomial  within  a 
parenthesis,  preceded  by  the  sign  — .  This  may  be  done 
without  altering  the  value  of  the  polynomial,  provided  the 
signs  of  all  the  terms  placed  within  the  parenthesis  are 
changed.  Thus,  a  —  m-\-n  may  be  written  a  —  {m  —  n); 
for  if  the  subtraction  indicated  in  the  latter  expression  be 
performed,  we  obtain  a  —  m-f-n. 

Let  the  student  throw  all,  except  the  first  term,  of  each 
of  the  following  polynomials  into  a  parenthesis,  preceded 
by  the  sign  — . 

1.  m^  —  c-^-a. 

2.  4a2-^6  — 3c2. 

3.  150  — 2x2  +  4y  — 7a6. 

4.  62:2^2_|.2ax  +  36y  — IO22. 

5.  a^m^—lO  +  Qxj/, 

6.  Sabc-\-mx  —  Si/-\'10. 

7.  4 x2 y3  —  6x1/^ -^abc  —  nfi x. 

8.  20j?2g  — 56  +  a;2. 

9.  a262c2+10  — a36c2  +  a5c. 


'^•*W«5Bi^5* 


64  MULTIPLICATION    OF    POLYNOMIALS.  [^H. 

SECTION  XI. 

MULTIPLICATION    OF    POLYNOMIALS. 

Art.  4:1.  Let  it  be  required  to  multiply  7 -f- 3  by  5. 
Since  7  +  ^  is  10,  the  product  must  be  5  .  10,  or  50. 
Now,  to  multiply  without  reducing  the  multiplicand,  it  is 
evident  that  we  must  multiply  both  7  and  3  by  5,  and  add 
the  products.  ^ 

Operation, 
7  +  3 

_5 

Product  =  35 +  15:=  50. 
In  like  manner,  to  multiply  a  +  6  by  c,  we  must  multi- 
ply both  a  and  h  by  c,  and  add  the  products. 
Operation, 
«  +  6 


Product  zz:  a  c  +  6  c. 
But  if  any  term  of  the  multiplicand  has  the  sign  — ,  the 
monomial  multiplier  being  affected  with  the  sign  +,  the 
corresponding  product  must  have  the  sign  — . 

As  an  example  in  numbers,  let  us  multiply  9  —  5  by  3. 
Since  9  —  5  is  4,  the  product  must  be  3.4,  or  12.  But 
if  we  first  multiply  9  by  3,  the  product  27  is  too  great  by 
3  times  5,  or  15 ;  we  must,  therefore,  multiply  5  by  3, 
and  subtract  the  product  15  from  27,  which  gives  27  — 
15,  or  12. 

Operation, 
9  —  5. 
J 

Product  =  27  — 15=12. 


»§>ll.|  MULTIPLICATION    OF    POLX^NOMIALS.  65 

In  a  similar  manner,  to  multiply  a  —  6  by  c,  if  we  first 
multiply  a  by  c,  the  product  a  c  is  too  great  by  c  times  h ; 
we  must,  therefore,  multiply  h  by  c,  and  subtract  the 
product  h  c  from  a  c,  which  gives  ac  —  he  for  the  correi  t. 
result. 

Operation, 

a  —  h 


Product  m  «  c  —  6  c. 

Hence  we  see  that  — 6,  multiplied  by  -j-  c,  gives  — he 
If  both  of  the  quantities  whose  product  is  sought  art 
polynomials,  the  whole  of  the  multiplicand  must  be  multi- 
plied by  each  term  of  the  multiplier. 

Let  it  be  required  to  multiply  7 -(-2  by  4-|-3.  Since 
7  4-2  is  9,  and  4  -|-  3  is  7,  the  product  must  be  7 . 9,  or 
63.  To  produce  this,  7  +  2  must  first  be  multiplied  by 
4,  which  gives  28 -f- 8;  then  7-i-2  must  be  multiplied  by 
3,  which  gives  21  -f-  6 ;  the  sum  of  these  products  is  28 
+  8  +  21-1-6,  or  63. 

Operation. 
7  +  2 
4  +  3 

Product  =  28  +  8  +  21  +  6  =  63. 

In  a  similar  manner,  to  multiply  a  +  6  by  c  +  c?,  we 
first  multiply  a  +  6  by  c,  which  gives  ac-\-hc\  then 
multiply  a  +  6  by  c?,  which  gives  ad-\-bd\  now,  addmg 
these  products,  we  have  «c  +  6c  +  ac?+6c?  for  the  en- 
tire product. 

Operation. 

a  +  6 
c^rd 

Product  =zac-\-hc-\-ad-\'hd, 
5 


66  MULTIPLICATION    OF    POLYNOMIALS.  [*§>  H- 

Now,  let  it  be  required  to  multiply  6  —  2  by  5  —  3. 
Since  6  —  2  is  4,  and  5  —  3  is  2,  the  product  must  be 
4 .  2,  or  8.  If  we  first  multiply  6  —  2  by  5,  we  have  30 
— 10,  or  20^;  this  is  too  great  by  3  times  6  —  2,  which  is 
18  —  6,  or  12 ;  we  must,  therefore,  subtract  18  —  6  from 
30  _  10,  which  gives  30  — 10  —  18  +  6,  or  8. 
Operation 
6  —  2 

o  —  t) 

Product  =  30  —  10  —  18  -f  6  zz:  8. 

In  like  manner,  to  multiply  a  —  b  by  c  —  d,  we  first 
multiply  a — 6  by  c,  and  the  product,  as  has  already  been 
seen,  is  ac  —  be;  this  is  too  great  by  d  times  a  —  6, 
which  is  ad  —  bd;   this  last  product  must  therefore  be 
subtracted  from  ae  —  6c,  which  gives  ac  —  be  —  ad-\' 
b  d,  for  the  true  product  of  a  —  6  by  c  —  d 
Operation, 
a  —  b 
c  —  d 


Product  =z  ae  —  be  —  a d -{-  b d. 
On  examining  the  preceding  product,  we  see  that  — ad 
was  produced  by  multiplying  -\-  a  by  —  d;  hence,  if  a 
term  having  the  sign  -\-  be  multiplied  by  a  term  having 
the  sign  — ,  the  corresponding  product  must  have  the 
sign  — .  Also,  -\-bd  was  produced  by  multiplying  —  b 
by  —  d;  hence,  if  a  term  having  the  sign  —  be  multi- 
plied by  another  term  having  the  sign  — ,  the  product 
must  have  the  sign  -\- ;  in  other  words,  if  two  negative 
terms  are  multiplied  together,  the  product  must  be  posi- 
tive. 

Art.   42.     From  the  preceding  analysis,  we  derive 
the  following 


•J*  1  1  .]  MULTIPLICATION    OF    POLYNOMIALS  67 

RULE    FOR    THE    MULTIPLICATION    OF    POLYNOMIALS. 

1.  Multiply  all  the  terms  of  the  multiplicand  hy  each 
term  of  the  multiplier  separately,  according  to  the  rule  for 
the  multiplication  of  monomials. 

2.  With  regard  to  the  signs,  observe,  that  if  the  two 
terms  to  he  multiplied  together  have  the  same  sign,  either 
both  -\-,  or  both  — ,  the  corresponding  product  must  have 
the  sign  -f- ;  but  if  one  term  has  the  sign  -f-,  cind  the 
other  the  sign  — ,  the  corresponding  product  must  have 
the  sign  — 

tj.    Add  together  the  several  partial  products,  reducing 
terms  which  are  similar, 
1     Multiply  ^ax-\-l  c^y 
by  2  a  a;  -f-  5  c^  y 

Sa^x^-^Uac^xy  \    Partial 

-f  20  a  c2  x  y  +  35  c^  y2  I  products. 

Sa^x^-\'Mac^xy-\-Z6c^y^',    the  entire 
product  reduced. 

To  facilitate  reduction,  it  is  advisable  to  place  the  simi- 
lar terms  of  the  partial  products  under  each  other. 

2.  Multiply 
2a26  +  36c  — c2   by 

a^h  —  ^bc 

2a'^b^-\-Sa^b^c  —  a^bc^  \    Partial 

_  5  «2  52  c  _  15  62  c2 -(- 5  J  c3  }  products. 

2a4  62_2a2  62c_a2  6c2— 15  62c2-f-5  6c3.     Result. 

3.  Multiply 

x^  +  x^y  +  x^y^-^xy^-^y^,  by 
x  —  y 

x^'\'X'^y-^x^y^-\'X^y^-\-xy'^  \    Partial 

—  x^y  —  x3  y2  —  x^y^  —  xy"^  —  y^  i  products. 

x5 y5      Result. 


68  MULTIPLICATION    OF    POLYNOMIALS.  [*§>  1 1 

4.  Multiply  7 ab-{-'3cd  by  2mn-\^4L  ex, 

5.  Multiply  15  x  y  +  7  a;2  —  4  m  by  3  2:  y  —  2x^. 

6.  Multiply  3 «2_2 a 6 4-5 62  by  3a  — 56. 

7.  Multiply  «2  _^  a  6  +  62  by  a  —  6. 

8.  Multiply  x4  —  3/4  by  x^  +  y4, 

9.  Multiply  2x2  — 3 xy  + 6  by  3x2-|-3xy  —  5. 
10.  Multiply  3a  — 2  6 +  2c  by2a  — 46  +  5c. 

Art.  43.  The  multiplication  of  polynomials  is  indi- 
cated by  putting  the  quantities  under  a  vinculum,  or  within 
parentheses,  and  writing  them  after  each  other,  with,  in  the 
former  case,  and  either  with  or  without  the  sign  of  multi- 
plication between  them,  in  the  latter.  It  is  most  usual  to 
omit  the  sign  of  multiplication.  Thus,  each  of  the  expres- 
sions, x  +  i/Xx  —  y,  x-^y.x  —  y,  (a: +  y)  X  (x  — y). 
(x-|-y).(x  —  y),  and  (x -{- y)  (x  —  y),  indicates  the 
multiplication  of  x-j-y  by  x  —  y. 

The  parenthesis  is  generally  preferable  to  the  vinculum  ; 
but  the  learner  must  be  careful  to  include  the  whole  of 
each  polynomial  within  the  parenthesis;  for  (a-f-6)  x  —  y 
signifies  that  a-f-6  is  multiplied  by  x  only,  and  that  y  is 
subtracted  from  the  product. 

The  expression  (a -|- 6)  (c  —  d)  (x-j-y)  indicates  that 
the  first  polynomial  is  to  be  multiplied  by  the  second,  and 
that  product  by  the  third. 

Let  the  student  oerform  the  operations  indicated  in  the 
following  examples. 

1.  (m2  4-w2)  (m^  —  n^), 

2.  (a2  +  2a6  +  62)  (a  — 6). 

3.  (4a  +  562_6c6;2)  (5a  — 762). 

4.  (3a2  — 5a6  +  2)  (4a2-fl0a6  — 6). 

5.  (x  +  y)(x-y)(x2  +  y2). 

6.  {a^b  —  Scd)  (a2  +  xy)  +  (3a2  6  +  3c(/)  ya^  — 
3xy). 

In  this  last  example,  the  product  of  the  third  and  fourth 


'^'ll.]  MULTIPLICATION    OF    POLYNOMIALS.  69 

polynomials  is  to  be  added  to  the  product  of  the  first  and 
second. 

7.  (Sm^  +  Smn)    {Sx  + 2  i/) -\- {5m^  +  4  ax)    (2  x 

8.  (17x2_2y)  (a24-26)  +  (25a:2  +  3y)  (3a2_ 
5  6). 

9.  (6x7/'^Saz)  {2x^  —  5az)  —  {2xy  —  mx)  {2x^ 
—  4tmx), 

In  the  9th  example,  the  product  of  the  last  two  quanti- 
ties is  to  be  subtracted  from  that  of  the  first  two. 

10.  (Sab-^cd)  (m^x  +  wy)  — (3m2x  — 4wy)  2ab. 
11     (a5  +  5a4a:  +  2y2)  b^  c  —  {a^ -^  4a^x)   {ab^c  — 

my2). 

12.    {5a^  —  Sab  +  4b^)  (6a  — 5  6)  — 3a  (a^  +  a^). 

Art.  4  J:.  The  following  examples  deserve  particular 
attention,  on  account  of  the  use  which  will  hereafter  be 
made  of  the  results. 

The  sum  of  the  two  quantities,  a  and  6,  is  a  -|-  6,  and 
their  difference  is  a  —  b.  Let  us  multiply  this  sum  bv 
this  difference. 

Operation, 
a  +  b 
a  —  b 


a^  +  ab 
—  ab  —  b^ 

a^  —  b^. 

This  product  is  the  difference  between  the  second 
power  of  a,  and  the  second  power  of  b.     Hence, 

The  sum  of  two  quantities  multiplied  by  their  difference  . 
gives  the  difference  of  the  second  powers  of  those  quantities 

According  to  this  principle,  (8 -f- 3)  (8  —  3)  =  64  — 9 
=  55,  and  (3<7  +  .r)  (3 a  —  x)  =  9 a^ _ 2.2.. 


TO  MULTIPLICATION    OF    POLYNOMIALS.  [•§>  1  1 - 

Let  the  learner  give  the  results  of  the  following  exam 
pies,  without  actually  multiplying. 

1.  (x  +  y)  (x  — y). 

2.  {m-\-n)  (m  —  n). 

3.  (3«  — 4a:)  (3a-t-4a;).        /.     ' 

4.  (4a2  +  3y)(4a2  — 3y).;^.^ 

5.  (10  +  2a)  (10  — 2a).      //-^^.4/V 

6.  (6  a2  6  _  3  m)  (6  a2  6  -f  3  m),  f  /  .^  ^  t  f^^.   ^ 

Art.  45.  When  a  polynomial  is  multiplied  by  itself, 
the  product  is  called  the  second  power,  and  when  multi- 
plied twice  by  itself,  the  product  is  called  the  third  power 
of  that  polynomial. 

Find  the  second  power  of  a  +  ^• 

Operation, 
a  +  6 
a  +  6 

a^-\-ab 
+  a6  +  62 


a2  -|-  2  a  6  +  62  =  2d  power  of  a  +  6. 

Hence  the  second  power  of  the  sum  of  two  quantities 
contains  the  second  power  of  the  first  quantity,  plus  twice 
the  product  of  the  first  by  the  second,  plus  the  second 
poioer  of  the  second. 

Find  the  second  power  of  a  —  h. 

Operation, 
a  —  h 
a  —  h 


a^  —  ah 
—  a6  +  62 


fl2_2 a 6 -f  62  =  2d  power  of  a  —  6. 


<§>ll.J  MULTIPLICATION    OF    POLYNOMIALS.  71 

This  result  differs  from  the  second  power  of  a-f-  ^  only 
in  the  sign  of  2  ab,  which  is  in  this  case  negative. 

By  means  of  the  two  principles  in  this  and  the  preced- 
ing article,  write  the  second  powers  of  the  following 
quantities. 

1.  x  +  y.^^X^^^T^  '  6.    7  m  — X.      ;/>^v%^v'^^>, 

2.  X  —  y.   y^  2:jL  4-'^      '^'    3a:  +  5y. 

3.  /w  +  w.  .r^iViKv'/-^^  '^-    a^  +  ^b^' 

4.  m  —  n.  9.    10  —  3x2. 

5.  2x  +  St/.^  10.   3xy  +  2/w2. 

Art.  4:6.  If  the  second  power  of  «  -f-  5  is  multiplied 
by  a-\-b,  the  product  will  be  the  third  power  of  a-\-b, 
which  is 

a^-]-Sa^b-^Sab^  +  b\ 

Hence  the  third  power  of  the  sum  of  two  quantities 
contains  the  third  power  of  the  first  quantity^  plus  three 
times  the  second  power  of  the  first  into  the  second,  plus 
three  times  the  first  into  the  second  power  of  the  second , 
plus  the  third  power  of  the  second. 

In  like  manner,  the  third  power  of  a  —  b  will  be  found 
to  be 

which  differs  from  that  of  «  +  6,  in  having  its  second  and 
fourth  terms  negative. 

Write  the  third  powers  of  the  following  quantities. 

1.  x  +  y.  5.   2a  +  7». 

2.  X  —  y,  6.    4tx — y, 

3.  m  +  n.  7.    3wi4-2«. 

4.  m  —  n  8.   2^2  — 3y2 


73  DIVISION    OF    MONOMIALS.  [*§»  12. 


SECTION   XII. 

DIVISION    OF    MONOMIALS. 

Art.  4:7.  Division  being  the  reverse  of  multiplica- 
tion, the  object  is  to  find,  for  the  quotient,  a  quantity, 
vv^hich,  when  multiplied  by  the  divisor,  shall  produce  the 
dividend :  in  other  vi^ords,  having  a  product  and  one  of  its 
factors,  the  object  is  to  find  the  other  factor. 

Since,  then,  the  product  of  the  divisor  and  quotient  must 
reproduce  the  dividend,  the  coefficient  of  the  quotient  must 
be  such,  that,  when  multiplied  by  the  coefficient  of  the  di- 
visor, it  shall  produce  that  of  the  dividend ;  and  the  expo- 
nent of  any  letter  in  the  quotient  must  be  such,  that,  when 
added  to  the  exponent  of  the  same  letter  in  the  divisor, 
it  shall  give  the  exponent  of  that  letter  in  the  dividend. 
Also,  it  is  manifest  that  the  quotient  must  contain  those 
letters  of  the  dividend  which  are  not  found  in  the  divisor. 

Divide  mahy  m,  or  find  —  of  ma.     The  answer  is  a  ; 

m 
because,  if  m  be  multiplied  by  a,  the  product  is  m  a. 


Divide  ax 

by  a. 

Ans. 

X. 

Divide  3«6 

by  36. 

Ans. 

a. 

Divide  6amx 

by  2  a, 

Ans. 

Smx. 

Divide  a,  or  I  a 

by  1. 

Ans. 

a. 

Divide  a,  or  la 

by  a. 

Ans. 

1. 

Divide  ab  c 

by  ab  c. 

Ans. 

1. 

Divide  7abx 

by  ax. 

Ans. 

7  6. 

Divide  15  am 

by  5  m. 

Ans. 

3  a. 

We  readily  see  the  correctness  of  the  preceding  an- 
swers, because,  in  each  case,  the  product  of  the  divisor 
and  quotient  gives  the  dividend. 


§  12.]  DIVISION    OF    MONOMIALS.  73 

Divide  a^  by  a^.  The  quotient  is  a^,  because  c^ .  c? 
=  aK 

Divide  la^h^c  by  1  ah^.  The  quotient  is  ah^Cy  be- 
cause 1  ah^ ,  aW cz=zl  a^Jr* c. 

We  found,  in  the  multiplication  of  monomials,  that,  when 
the  same  letter  occurred  in  both  multiplicand  and  multi- 
plier, the  exponents  were  added,  to  obtain  the  exponent  of 
that  letter  in  the  product.  On  the  other  hand,  in  dividing, 
as  we  see  in  the  last  two  examples,  the  exponent  of  any 
letter  in  the  divisor  is  subtracted  from  the  exponent  of  the 
same  letter  in  the  dividend,  in  order  to  obtain  the  expo- 
nent of  that  letter  in  the  quotient.  Thus,  a^,  divided  by 
a^,  gives  a^~^=z  a^. 

Art.  48.  From  what  precedes,  we  deduce  the  fol- 
lowing 

RULE    FOR    DIVIDING    ONE    MONOMIAL    BY    ANOTHER. 

1.  Divide  the  coefficient  of  the  dividend  hy  the  coeffi- 
cient of  the  divisor. 

2.  Strike  out  from  the  dividend  the  letters  common  to 
it  and  the  divisor,  when  they  have  the  same  exponents  in 
both ;  hut  if  the  exponents  of  any  letter  are  different ^  sub- 
tract its  exponent  in  the  divisor  from  its  exponent  in  the 
dividend,  and  write  the  letter  in  the  quotient  with  the  re* 
mainder  for  an  exponent. 

3.  Write  also  in  the  quotient,  with  their  respective 
exponents,  the  letters  of  the  dividend  not  found  in  the 
divisor. 

Remark.  If,  however,  the  divisor  and  dividend  are,  in 
any  case,  alike,  the  quotient  is  1. 

1.   Divide  6a^b^   '  by  2  a  6.         Ans.  3a62. 
2     Divide  25  a^         by  5  a^.  ^^ 


74 


DIVISION    OF    POLYNOMIALS. 


[§\3 


3. 
4. 
5. 

6. 

7. 

8. 

9 

10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 
21. 
22. 
23. 


16  x^  1/ 

17  am^x^ 
21x^7j^z 
a^x^ 
S5x'^t/^ 

64  a^  m^  x 
SSx^fz^ 
150am^x'^ 

17  ax 

18  a^b^c^x 
2Smxp^ 
SSSa^x7/^z 
lllp^x^y^ 
99  a^ 

17  a  m^  a:  4 
3  a^  b^  c^  2:7 
75mi0y9 
125x8^7  2:6 

3«26c 
lOOmiOxyS 


by  2x2y.  ^x^ 
by  ax. 


by  3...      ^,y 


by  a2 

by  7xy3.    ^Tia     -i 
by  10m3y3jJ-^*y^ 

by  a^m^x.  Ci^df  dw 
by  11x^2^3^  J  y\^  '7 
by  30ax2.     i*^  ;^ 
by  ax.  Ij       . 

by  6a62c.     io^'^ 
by  4 m x'p,     7'p^ 
by  9axy5.   Jn^^ 
by  3p2y.     ^f^f^^^- 

by  ax^.     /y  tyyv  ", 

by  3a62c3    ^:iu6r  x^      j 

by  25mTi/\  Son.^^ 


Y 


y 


r^  ^' 


by  5x3  3^2^,4^^^  J/    J 
by  4w3xv^.  t?/">^t  ^ 


SECTION   XIII. 


DIVISION    OF    POL11NOM1AL8. 

Art.  49,  If  a-\-m  —  c  be  multiplied  by  d,  the 
product  will  be  ad-^dm  —  cd;  consequently,  if  a tZ  + 
dm  —  c  c?  be  divided  by  d^  the  quotient  will  be  a-\-m  —  c. 
This  quotient  is  obtained  by  dividing  each  term  of  the 
dividend,  ad-^dm  —  c  d^  by  the  divisor   d.     In   order, 


^  13.]  DIVISION    OF    POLYNOMIALS.  75 

therefore,  that  a  polynomial  may  be  divisible  by  a  mono- 
mial, each  term  of  the  dividend  must  be  divisible  by  the 
divisor.  The  rule  for  the  signs  of  the  partial  quotients 
may  be  determined  as  follows,  if  we  recollect  that,  in 
all  cases,  the  product  of  the  divisor  and  quotient  must 
give  the  dividend. 

If  -\-am  be  divided  by  -\-  «,  the  quotient  must  be  -f-  m, 
because  the  product  of  -j- a  and  -\-m  is  -\- am. 

If -|-«m  be  divided  by  —  «,  the  quotient  must  be  — m, 
because  the  product  of  —  a  and  —  m  is  -\- am. 

If  —  am  be  divided  by  -|-«,  the  quotient  must  be  — w, 
Decause  the  product  of  -f-  a  and  — m  is  — am. 

Finally,  if  — am  be  divided  by  — a,  the  quotient 
must  be  +  m,  because  the  product  of  — a  and  -^7n  is 
—  am. 

We  perceive,  therefore,  that  when  two  terms,  whose 
quotient  is  sought,  have  the  same  sign,  whether  both  -}- 
or  both  — ,  the  quotient  must  have  the  sign  -\- ;  but  when 
the  signs  of  the  two  terms  are  different,  the  quotient  must 
have  the  sign  — . 

Art.  50.  From  what  precedes,  we  derive  the  fol- 
lowing 

RULE     FOR     THE     DIVISION     OF     A     POLYNOMIAL     BY     A 
MONOMIAL. 

1.  Divide  each  term  of  the  dividend  hy  the  divisor,  ac- 
cording to  the  rule  for  dividing  one  monomial  hy  another , 
the  partial  quotients,  taken  together,  will  form  the  entire 
quotient, 

2.  With  regard  to  the  signs,  observe  that  when  the  term 
of  the  dividend  and  the  divisor  have  the  same  sign,  the 
corresponding  term  of  the  quotient  has  the  sign  -f- ;  but 
when  the  term  of  the  dividend  and  the  divisor  have  dif 


/6  DIVISION    OF    POLYNOMIALS.  [<§>  13 

ferent  signs ,  the  corresponding  term  of  the  quotient  has 
the  sign  — . 

1  Divide  am-\-7n^x  by  m.  Ans    a-^-mx. 

2.  Divide  3 x2y  + 6 2:  by  3a:. 

3.  Divide  I5m2/2_30y3  by  ^  f  ^/J  i^p^  ^  Q>  y 

4.  Divide  45  m^  —  15  m^  y  -j-  5  m  by  .  5  m.  ^ktC^^uiy 

-       5.  Divide  10^xy+16^2_|_4^^2  by  2 6?.  h^'^u  i^ tds^'^ 

6.  Divide   1 1  x^ y2  _j_  33  ^^ ^3  _|_ 22  2: y  by  \\  x yj y  f -^-^ (J 

7.  Divide  S4.  am^-{- 51  am^n— 17 a^m^  by  17a3|/._     / 
a  Divide  4i9x  —  63x7/^  —  56x^  by  7x.  -  v^^t,!,  0  ^^ 

-^.  Divide  —40x2  +  50x3/  — 30^3/2  by  —5x, 

10.  Divide  li) ab  c—l6  a^b^  —  20  a^b  by  — 2a&. 

11.  Divide  —  16m  x2  +  32  7;i2x  — 24m3x2  by  Smx. 

12.  Divide  —  45  a^  ^3  +  3  a^  x^ — 60  a^  ^2  by  —  3  a2  x2. 

•^">'    -  .-   iL  .^  ^  r.  ^^  /•/  V-  V  Z.::^  .--  -/  /^  ^C 
Art.  51,     When  both  dividend  and  divisor  are  poly- 
nomials, the  process  of  dividing  will  be  easily  understood, 
if  we  observe  the  formation  of  a  product  in  multiplication. 
Let  us  multiply  3a2  4-2a6by4a  +  3  6. 
Operation. 

3  a2  +  2  a  6 
4a  +  36 

12  a3  +  8  a2  ^ 

4-9  a2  6  +  6  a  62 


12  a3  +  17  a2  6  +  6  «  b^. 

If  this  product  be  divided  by  the  multiplicand,  the  quo- 
tient vi^ill  be  the  multiplier ;  or  if  it  be  divided  by  the  mul- 
tiplier, the  quotient  will  be  the  multiplicand. 

Since,  in  multiplying,  the  entire  multiplicand  is  multi- 
plied by  each  term  of  the  multiplier,  the  product,  if  no 
reduction  took  place,  would  contain  a  number  of  terms 
equal  to  the  product  obtained  by  multiplying  the  numbei 


I 


•^  vs.]  DIVISION    OF    POLYNOiVHALS.  77 

of  terms  in  the  multiplicand  by  the  number  of  terms  in 
the  multiplier.  Thus,  in  the  preceding  example,  there 
are  2  terms  in  the  multiplicand,  and  2  in  the  multiplier, 
and,  if  8  a^  6  and  9  (Pb  had  not  been  united,  there  would 
have  been  2.2  or  4  terms  in  the  product.  In  like  man- 
ner, if  one  factor  had  5  terms  and  the  other  3,  the  product, 
without  reduction,  would  contain  5 . 3  or  15  terms. 

But  generally,  by  reduction,  some  terms  of  the  product 
are  united,  and  others  are  cancelled  and  disappear.  There 
are,  however,  two  terms  of  the  product  which  can  neither 
be  united  with  others,  nor  disappear.  These  are,  1st,  the 
product  of  the  term  containing  the  highest  power  of  any 
letter  in  the  multiplicand  by  the  term  containing  the 
highest  power  of  the  same  letter  in  the  multiplier ;  2d,  the 
product  of  the  two  terms  containing  the  lowest  powers  of 
the  same  letter. 

Since  the  dividend  is  to  be  considered  as  the  product 
of  the  divisor  and  quotient,  it  is  plain  that  if  the  term 
containing  the  highest  power  of  any  letter  in  the  dividend 
be  divided  by  the  term  containing  the  highest  power  of 
the  same  letter  in  the  divisor,  the  result  will  be  the  term 
containing  the  highest  power  of  that  letter  in  the  quotient. 

Let  us  now  take  the  product  of  the  preceding  multipli 
cation  for  a  dividend,  and  the  multiplicand  for  a  divisor, 
and  see  how  we  can  obtain  the  multiplier,  considered  as 
a  quotient. 
V  1/^  Operation. 

Dividend.  Divisor. 

12a^  +  17a^b-{-6ab^  (^a^-\'2ab 


t 


12a3  +  8a26 (    4a  +  3  6.     auotient. 

9  a2  6  +  6  a  62 

9  a^  6  -j-  6  g  62 

0. 


78  DIVISION    OF    POLYNOMIALS.  [^,13 

According  to  what  has  been  said  above,  if  we  divide 
12  a^  by  3  a^,  we  shall  obtain  the  term  of  the  quotient 
with  the  highest  power  of  a.  This  division  gives  4  a  for 
the  first  term  of  the  quotient.  Then,  since  the  entire 
dividend  is  produced  by  multiplying  the  whole  divisor  by 
each  term  of  the  quotient,  if  we  multiply  the  whole  divisor 
by  4  «,  this  first  term  of  the  quotient,  the  product  will  be 
a  part  of  the  dividend.  This  product  is  \^n^  \-S  a^h, 
which  we  subtract  from  the  dividend,  and  find  for  a  re- 
mainder 9a^b-[-Qah'^, 

This  remainder  is  to  be  considered- as  a  new  dividend, 
and  as  produced  by  multiplying  the  divisor  by  the  remain- 
ing part  of  the  quotient ;  and  if  the  term  containing  the 
highest  power  of  a  in  this  new  dividend  be  divided  by  the 
term  containing  the  highest  power  of  a  in  the  divisor,  the 
result  must  be  a  new  term  of  the  quotient. 

The  division  of  9  «2  6  by  3  a^  gives  -f-  3  6,  which  we 
write  as  the  second  term  of  the  quotient.  We  now  mul- 
tiply the  whole  divisor  by  3  6,  and  the  product  must  be 
the  whole  or  a  part  of  the  new  dividend.  This  product 
is  9  «2  6  -(-  6  a  6^,  which,  subtracted  from  the  new  divi- 
dend, leaves  no  remainder.  The  entire  quotient,  there- 
fore, is  4  a  4~  ^  ^• 

Since  we  always  divide  the  term  containing  the  highest 
power  of  some  letter  in  the  dividend  by  the  term  contain- 
ing the  highest  power  of  the  same  letter  in  the  divisor,  it 
is  convenient  to  write  the  quantities  so  that  the  former  of 
these  terms  shall  stand  first  in  the  dividend,  and  the  latter 
first  in  the  divisor.  This  object  will  be  attained  by  ar- 
ranging both  dividend  and  divisor  according  to  the  powers 
of  the  same  letter. 

Remark.  A  polynomial  is  said  to  be  arranged  accord- 
ing to  the  powers  of  a  particular  letter,  when,  in  the 
successive  terms,  the  powers  of  that  letter  increase  or 


^  13.]  DIVISION    OF    POLYNOMIALS.  79 

diminish  from  left  to  right.  In  the  example  of  division 
just  given,  the  polynomials  were  arranged  according  to 
the  decreasing  powers  of  a.  The  same  arrangement 
should  be  preserved  in  each  partial  dividend,  as  was  made 
at  fir^t  in  the  entire  dividend. 

The  rule  for  the  signs  of  the  partial  quotients  is 
manifestly  the  same  as  that  given  for  the  division  of  a 
polynomial  by  a  monomial. 

Art.  53.  From  the  preceding  explanations,  we  de- 
duce the  following 

RULE    FOR    DIVIDING    ONE    POLYNOMIAL    BY    ANOTHER. 

1.  Arrange  the  dividend  and  divisor  according  to  the 
powers  of  the  same  letter,  beginning  with  the  highest. 

2.  Divide  the  first  term  of  the  dividend  by  the  first 
term  of  the  divisor,  and  place  the  result  as  the  first  term 
of  the  quotient ;  recollecting,  that  if  both  terms  have  the 
same  sign,  the  'partial  quotient  must  hojve  the  sign  4",  but 
if  they  have  different  signs,  the  partial  quotient  must  have 
the  sign  — . 

3.  Multiply  the  whole  divisor  by  this  term  of  the  quo- 
tient, subtract  the  product  from  the  dividend,  and  the 
remainder  will  form  a  new  dividend, 

4.  Divide  the  first  term  of  the  new  dividend  by  the 
first  term  of  the  divisor,  and  the  result  will  form  the 
second  term  of  the  quotient ;  multiply  the  whole  divisor  by 
this  second  term  of  the  quotient,  and  subtract  the  product 
from  the  second  dividend.  The  remainder  will  form  a 
new  dividend  from  which  another  term  of  the  quotient  may 
be  found. 

These  operations  are  to  be  repeated,  until  all  the  terms 
of  the  original  dividend  are  exhausted. 


80  DIVISION    OF    POLYNOMIALS.  [<§>  13. 

As  an  example,  let  us  divide  50  c^  6^  —  W  a^h-\-  20  a* 
~33a263_^  10^64  by  bah^ -\-ba^ —  ^a^h. 

Operation. 

20a^-16a^6  +  20a36S \  ^^^_^^j,^^ 

—25a%  +  30a362_33«253  ^  i  Oa^^ 
—25M  +  20^36^— 25a^63 

10a36'^— 8a263-|>l0a64 

0. 

First,  arranging  the  two  quantities  according  to  the 
powers  of  a,  we  place  the  divisor  on  the  right  of  the  divi- 
dend, separating  it  from  the  dividend  by  some  mark,  and 
draw  a  line  below  to  separate  it  from  the  quotient. 

We  now  divide  20  a^  by  5  a^,  and  have  for  a  quotienf 
-f-  4  a^,  which  we  write  as  the  first  term  of  the  quotient 
Multiplying  the  whole  divisor  by  4  a^,  we  place  the  prod- 
uct, 20tt5  — 16  «46+ 20^3  62,  under  the  dividend,  and 
subtract  it  from  the  dividend.  The  subtraction  is  per- 
formed by  changing  the  signs  of  the  terms  in  the  quantit}! 
to  be  subtracted,  considering  it  as  written  after  the  divi- 
dend,- and  then  reducing.  Thus,  changing  the  signs,  we 
have  —  20^5  _^  16  «4  6  _  20^3  62  .  then  +20^5  and 
—  20  a^  cancel,  —  41  a^  6  and  +  16  a^  b  make  — 25  a^  b, 
and  +  50  «3  52  and  —  20  a^  b^  make  +  30  «3  b^  ;  bringing 
down  the  remaining  terms  of  the  dividend,  we  have  for  a 
remainder  — 25  a^  6  +  30  a^  b^  —  33  a^  6^  + 10  «  6^  which 
we  regard  as  a  new  dividend. 

We  now  divide  — 25  a^  6  by  5a3^  and  obtain — 5ab, 
which  we  write  as  the  second  term  of  the  quotient.  Mul- 
tiplying the  whole  divisor  by  — 5ab,  we  subtract  the 
prodict,  —  25 a^ b  +  20 a^ b^  —  25 a^ b^,  from  the  second 


<^  13. J  DIVISION    OF    POLYNOMIALS.  81 

dividend,  and  find   for  a  remainder    10  a^b^  —  8a^b^-{- 
10  a  M,  which  forms  the  third  dividend. 

Dividing  10  a^b'^  by  5a^,  we  obtain  +2&^  which  we 
write  as  the  third  term  of  the  quotient.  Multiplying  and 
subtracting  as  before,  we  have  no  remainder.  The  entire 
quotient,  therefore,  is  4  a^  —  5ab-\-2b^,  the  correctness 
of  which  may  be  shown  by  multiplying  it  by  the  divisor, 
and  finding  that  the  product  is  the  same  as  the  dividend 

As  another  example,  divide  x^  —  3/^  by  x — y. 
Operation, 
x^  —  y^     C  X  —  y  


x^  —  x^yl^. 


x^  —  X'^y  i  ^3^^^y^xy^  +  y^ 


;3y_a;2y2 

■i-x^y^  —  y^ 

_j_2;2y2  —  xy^ 

-{-xy^- 
-\-xy3- 

-y* 
-y* 

0. 

In  this  example  several  terms  are  produced  in  the  pro- 
cess, which  are  not  found  in  the  dividend.  These  terms 
disappear  when  the  divisor  and  quotient  are  multiplied 
together. 

1.  J}W\dQa?b^-^bc^d-\-ac^d'\-a^b'^hy  a4-b. 

2.  Divide  9  a^  M  _|_  27  a3  ^2  ^2 -|- 18  a^  ^4   by   \\ab'^-\- 

3.  Divide  6  64 -[-12  62  a; -j-.  6  2:2  by  52-^2;. 

.  4.   Divide  6a3-i-563  +  23a26+22a62  by  2a  +  56. 
'  '    5.    Divide  x^  —  ^x^y^lOx'^if—lOx^y^-\-^xy'^-^ 
y^  by  2;2  —  2  a:  y  -|~  V'^- 

6.  Divide  16  x^  —  25  y^  by  4  2:2  +  5  y'\ 

7.  Divide  a^  —  \^a^v:^-\-Mx^  by  a^  —  iax-^ix^. 


82  DIVISION    OF    POLYNOMIALS.  [^  I  *J 

8.  Divide  6ax^—nabx  +  Sab^  by  Sx  —  b 

9.  Divide  10 a^  +  51  a^b^  — ^8 a^b  +  Aab^ ---ISb^ 
by  4  «  6  —  5  a2 -1-3  62. 

10.  Divide  Sx"^ -]'4:X^  i/  —  4.x^  —  4Lx^y^-^16xy  —  l5 
by  2a:y  +  2:2  — 3. 

Art.  59.  The  following  principles  in  division  will 
often  be  useful,  and  are  found  demonstrated  in  works 
designed  for  the  more  advanced  student. 

The  difference  between  similar  powers  of  two  quantities, 
the  exponent  of  the  powers  being  integral  and  positive,  is 
divisible  by  the  difference  of  those  quantities. 

Thus,  X  —  y,  x^  —  y^,  x^  —  y^,  x^ — y^,  &.C.,  are  each 
divisible  by  x  —  y,  :j    \r'y^fJVJ\/A^ 

Also,  the  difference  of  similar  eten  powers,  and  the  sum 
of  similar  odd  powers  of  two  quantities,  are  each  divisible 
by  the  sum  of  those  quantities. 

Thus,  a^  —  b^,  a'^  —  ^,  a^  —  W,  &c. ;  also,  a-\-b,  a^ 
-j-  b'^,  a^  -j-  b^,  &c.,  are  each  divisible  by  a-^-b. 


J 


1. 

Divide  x^  —  y^ 

by  X  —  y. 

2. 

Divide  x^  —  y^ 

by  x  +  y. 

3. 

Divide  x^-j-m^ 

by  %-^m. 

4. 

Divide  1 — m^ 

by  l-{-m. 

5. 

Divide  m^-\-n^ 

by  /w-f-w. 

Art.  54:.  When  a  product  is  represented  in  its  faxj- 
tors,  it  is  manifestly  divided  by  dividing  one  of  its  factors. 
Thus,  to  divide  5.9  by  3,  we  have  to  divide  the  9  only, 
and  the  quotient  is  5  .  3,  or  15.  In  like  manner,  to  divide 
15 .  16  by  12,  or  3.4,  we  divide  15  by  3  and  16  by  4, 
and  the  quotient  is  5.4,  or  20. 

Also,  to  divid-e  {m^  —  ^^)(^"t~y)  ^7  ^i  +  w,  we  divide 
the  factpr  m^  —  n^,  and  the  quotient  is  {m  —  w)(a:-|-y)- 
likewise    15  m3  («2  -|_  2  «  6  +  b'^)  (x^  —  y^)    divided    by 


<^  14.]  MULTIPLICAIION    OF    FRACTIONS.  83 

Sm{a-\'b)  {x  —  y ),  gives  5m^  {a-\-  b)  {x  -{-  y),  to  obtain 
which  we  divide  15  m^  by  3  m,  a^  -j_  2  a  6  +  ^^  ^^y  «  +  ^ 
and  x^  —  y^  by  x  —  y. 

1.  Divide  m^{x-^y)  by  wi^.  Ans.  m  (x-f-y). 

2.  Divide  a^  {m-^-n)  by  m  -f-  n,        Ans.  a^. 

3  Divide  4x5 (a +  6)  by  2x3.       ^^yC^l^^'i 

4.  Divide  Ibaim^-^-n^)  by  5 a.  ^^       X/^/zt  x  ^ 

/5.  Divide  21/^2  (^'  —  3/')  by3(x  +  y).")^^.'\/^^^^^^^ 

6.  Divide  3(^3— 1)  (a  — 6)  by  {m  —  V){a  —  b).     -- 

7.  Divide  16  m4  (a2  _  2  a  6  +  6^)   (c^  _  d^\  by 
4m3(a  — 6)(c  +  d).    ^.;;/  ^  ^)(Q-(?Q 

8.  Divide  6(m3+l)  (a3  — 1)  by  2(m+l)  {a— I) 


SECTION   XIV. 


MULTIPLICATION    OF   FRACTIONS    BY    INTEGRAL    QUANTITIES 

Art.  55.     Fractions  have  the  same  meaning  in  alge- 
bra that  they  have  in  arithmetic.     Thus,  —  signifies  that 

n 
a  unit  is  divided  into  n  equal  parts,  and  that  m  of  those 
parts  are  taken  ;  or  it  expresses  division,  and  signifies  that 
m  is  divided  into  n  equal  parts. 

1.  How  much  is  3  times  f  1     Ans.  f . 

2.  How  much  is  5  times  —  1     Ans.  — . 

n  n 

3.  How  much  is  c  times  —  1     Ans.  — . 

6  6 

4.  What  is  I  of  5  ?     ^  of  5  is  |,  and  J  of  5  is  %\  \ns 

5.  What  is  f  of  a  ?     ^  of  a  is  — ,  and  f  of  a  is  — ,  Ans 

5  5 


84  MULTIPLICATION    OF    FRACTIONS.  [<^  14 

6.    What  is  the  —  part  of  a  ?    —  of  a  is  —,  and  —  of  a 

n  n  n  n 

am      . 
18  — ,  Ans. 

n  ' 

In  the  first  three  of  the  preceding  questions,  the  object 
was  to  multiply  a  fraction  by  a  whole  number ;  and  in  the 
last  three,  to  find  a  fractional  part  of  a  whole  number, 
that  is,  to  multiply  a  whole  number  by  a  fraction ;  and  we 
perceive  that  in  both  cases  we  multiplied  the  numerator 
and  the  whole  number  together. 

Hence,  to  multiply  a  fraction  hy  an  integral  quantity, 
or  an  integral  quantity  hy  a  fraction,  multiply  the  nu- 
merator by  the  integral  quantity,  and  write  the  product 
over  the  denominator. 


1. 

Multiply  -— 

by  m,     Ans.  — . 

3. 

Multiply  ""^y 

,               .          cx-^-cy 
by  c.     Ans.    — - — . 

3. 

Multiply 

x  +  y 

by  a. 

4. 

Multiply  ^^^ 

by  m. 

5. 

Multiply  '^+^ 

xy 

by  6. 

6. 

Multiply  — 

byx2  +  y2.       - 

7. 

Multiply  ^- 

by  a- — h. 

8. 

Multiply  """^^ 

hy  m^  —  n^. 

9. 

What  is  the  ^ 

part  of  2a;  +  3y; 

i'»- 

What  is  the  ^ 

m 

-  part  of  2  c  —  xl 

u. 

Multiply  

x-^y 

hy  2c  +  d,    .      ^d 

^ 


^  14. J  MULTIPLICATION    OF    FRACTIONS.  85 

13.    Multiply  -  by  3.  mx 

^  L 

The  fraction   —  means  that  m  is  divided  into  6  euual 

6 

parts.  If  we  divide  the  denominator  by  3,  which  gives 
— ,  m  is  then  divided  into  J  as  many  parts  as  it  was  before ; 
consequently,  the  parts  are  3  times  as  great  as  they  were 
before ;  that  is,  3  times  —  is  — . 

'  '  6  2 


14.    Multiply  —  by  m. 


a 
mn 


a 


The  fraction  —  means  that  a  is  divided  into  m  n  equal 


mn 


parts.     If  we  divide  the  denominator  by  w,  which  gives 
— ,  a  is  then  divided  into  —  as  many  parts  as  it  previously 

71  771 

was;   the  parts,  therefore,  are  m  times  as  great  as  they 
were  before ;  that  is,  m  times  —  is  — . 

77171  71 

Hence,  to  multiply  a  fraction  and  an  integral  quantity 
together,  divide  the  denominator  hy  the  integral  quantity, 
if  possible. 

Art.  5G«  Combining  this  rule  with  the  preceding, 
we  have  a 

GENERAL     RULE    TO     MULTIPLY    A    FRACTION    AND    AN 
INTEGRAL    QUANTITY    TOGETHER. 

Divide  the  denominator  hy  the  integral  quantity,  if 
possible ;  if  not,  multiply  the  numerator  by  the  integral 
quantity. 


86 


MULTIPLICATION    OF    FRACTIONS. 


[<^14. 


The  learner  may  perform  the  following  examples  by 
dividing  the  denominator. 


1.    Multiply   ^ 


2.  Multiply 

3.  Multiply 

4.  Multiply 

5.  Multiply 


a  +  36      ' 

4x2  1/3 

am  -\-  en 

Wjc^z^ 

ax^y 

X2-2/2 

4x— 171/ 

a2+2a6-|-62 


d 


\  6.   Multiply  ^!5l±l^ 
>^     7.    Multiply  a^^ab  +  h^ 


^ //■     8.   Multiply  x^  +  y2 

-  9.    Multiply  3  (m  +  «)  («  — 6)  by '^^^-^^^V (^  y  ■ 

10.    Multiply  4y2(c,  +  y)  by      (^ « +  6)  ("» + »)_ 


11.    Multiply 


by  6. 

a 


Dividing  the  denominator,  we  have  — ,  or  a. 

12.    Multiply^  by  m^n. 

Dividing  the  denominator  by  rw^  n,  we  have 
a  +  b.     Hence, 


a+b 


or 


If  a  fraction  be  multiplied  by  a  quantity  equal  to  iti 
denominator^  the  product  will  be  its  numerator. 


13    Multiply  ?^ 


by  4  6. 


14. 

Multiply  "+* 

15. 

Multiply ;;; 

16. 

Multiply  — — 

^^    3a{x  +  y) 

17. 

Multiply ^ 

^ -^        m2  — n2 

18. 

n*    1,-    1      4a6c  — xy 

19. 

Multiply  

<5>  14.]  MULTIFI*»C?ATJON    OF    FRACTIONS.  81' 

by  a  x^. 

by  a  +  ft. 

by  3a(x  +  y). 

by  (m  +  w)  {m  —  n). 

by  3acx  —  Sadx 

by  2x2y  +  2a;y2. 

'^**'2/  V*  T"  y/ 

Some  of  the  following  examples  can  be  performed  by 
dividing  the  denominator ;  in  others,  the  numerator  must 
be  multiplied. 

20.    Multiply  — —  by  2x^y.    ^^ 

2L  Multiply  -1±^  by  a +  6.  •- -^^ 

M"^*iply4-^^  by  2(^2  +  ^2).    .^^^--Ot'' 

23.  Multiply  3(2:2_y2)  by  _-^l_.         -rrzp^%\ 

^.  Multiply  a6m2  by  ^-^^.  0-^i'- A  'f 

25.  Multiply  x-\-y 

26.  Multiply  a  +  6 

27.  Multiply  x^  +  yS 

28.  Multiply  x^  —  2xy  +  y^hy 

29.  Multiply  ^-^^^^^  by  4(a  +  x).        ^'     ' 

30.  Multiply  3(a;4-y)  by      """^ 


UJ 

12(a:4-.y4) 

by 

by 

m* 

ax  A-av 

by 

a  +  6 
3x2/* 

hv 

ofi^y2 

a+b 


V^r-'-^ 


di. 


•"•-'   i^^ 


88  MVISION    OF    FRACTIONS.  [^  15. 

SECTION   XV 

DIVISION    OF    FRACTIONS    BY    INTEGRAL    QUANTITIES. 

Art.  57.     Divide  f  by  2,  or  find  J  of  f .     Ans.  f . 
Divide  —  by  3,  or  find  4  of  — .     Ans.  — . 

Divide  —  by  a,  or  find  —  of  — .     Ans.  — , 

m  a  m  m 

These  answers  may  be  shown  to  be  correct  by  the  fact, 
that  the  quotient  multiplied  by  the  divisor  produces  the 
dividend. 

Hence,  to  divide  a  fraction  hy  an  integral  quantity^ 
divide  the  numerator  hy  the  integral  quantity,  if  possible^ 

I.   Divide  — 
6c 


Ai«. 

a+6 

3. 

4. 

Divide  '""  +  '^ 
5c 

5. 

Divide  ^^  +  ^^^ 

a-\-b 

f\ 

T^.    .,      3a26  +  9a362- 

-21a* 

7.    Divide 


d?'-b^ 


8.  Divide  '-i^+^> 

6{a  — 6) 
^     -rx.    .,      16a2  +  32ax+16x2 

9.  Divide  

15(rr  — y) 


by  3w. 

A         2m 
Ans.  -— . 

by  5x^y. 

by  5a2c. 

ifa-ii-^ 

by  a. 

t-n.-h'l' 

yc 

by  3  a;. 

'■^l  . 

by  3a2. 

^^*/t^^ 

by  a  —  b. 

by  x-\-y. 

by  a2  +  2 

> 


<§>  15.]  DIVISION    OF    FRACTIONS.  89 

10.  Divide  L^^  +  ^^--±i^-      by4x  +  7    ^^ 

11.  Divide  —  by  a 

In  this  example,  the  numerator  cannot  be  divided  by  the 
divisor.  But  we  have  seen  that  a  fraction  is  multiplied 
by  dividing  the  denominator ;  on  the  other  hand,  a  frac- 
tion is  divided  by  multiplying  the  denominator. 

The  fraction  —  means  that  m  is  divided  into  n  equal 

n 

parts ;  and  if  the  denominator  be  multiplied  by  5,  for  ex- 
ample, m  will  be  divided  into  5  times  as  many  parts  as  it 
was  before  ;  consequently,  the  parts  will  be  -J  as  great  as 

before,  that  is,  4  of  —  is  — .     In  like  .manner,  if  the  de- 

n  5  n 

nominator  of  —  be  multiplied  by  a,  m  will  be  divided  into 
n 

a  times  as  many  parts  as  it  was  before,  aud  the  parts  will 
be  —  as  great  as  before ;  that  is,  —  of  —  is  — . 

a  a  n         an 

Hence,  to  divide  a  fraction  hy  an  integral  quantity , 
multiply  the  denominator  hy  the  integral  quantity. 

Art.  58.  Combining  this  rule  with  the  preceding, 
we  have  the  following 

GENERAL    RULE     FOR    DIVIDING    A    FRACTION    BY    AN    INTE- 
GRAL   QUANTITY. 

Divide  the  numerator,  if  it  can  he  done,  if  not,  multiply 
the  denominator,  hy  the  integral  quantity. 

1.    Divide  by  2ac^.  Ans. 


2.    Divide by  Qah.  Ans.  -. 

Sbc  ^  18a6»c 


m 


FACTORS. 


3.  Divide 

46  — c 

4.  Divide  

X  — y 

5.  Divide  ?<^±i) 

Imn 


by  5  m^  y 

by  x-f  y. 
by  a-|-x, 


6.   Divide  ^"'^  +  ^''^  +  ^'>    by  «  +  6, 
9(1  +  3/)  ' 


7.  Divide  ^=^ 

8.  Divide  ii^^±^^ 

9.  Divide 


by  x  —  y 


i 


[^16. 


^&^'^K/>f(yn  fK^  oy 


10.   Divide 


Mabc 

3(2^2-^2) 


a+6 


by  x^-^y^. 
by  wi-f-w. 


11.   Divide  <°'-^j^'^'-^'    by  («  +  6)(.-y) 


12.   Divide  fi^±J^ 

X3  — t/3 


by  5  (a:^  +  y^)- 


<^>'^*^^^ 


SECTION  XVI. 


^■^ 


FACTORS,    OR    DIVISORS    OF    ALGEBRAIC    QUANTITIES. 

I  Art.  59.  A  prime  quantity  is  one  that  can  be  divided 
by  no  entire  and  rational  quantity,  except  itself  and  unity. 
Thus,  a,  6,  and  a-\-m  are  prime  quantities. 

Two  quantities  are  prime  with  regard  to  each  other, 
ivhen  no  quantity,  except  unity,  will  divide  them  both 
without  a  remainder.  Thus,  a  h  and  c  d,  although  neither 
of  them  is  a  prime  quantity,  are  prime  with  regard  to 
each  ether. 


§>  16.]  j/iPACTOlis)  ftl 


Remark.  Although,  algebraically  considered,  we  call 
«,  h,  and  c,  prime  quantities,  they  are  strictly  speaking 
such,  only  when  they  represent  prime  numbers. 

We  have  frequent  occasion  to  separate  quantities  into 
their  prime  factors.  In  a  monomial,  this  operation  is 
attended  with  no  difficulty.  We  have  only  to  find,  ac- 
cording to  the  method  usually  given  in  arithmetic,  the 
prime  factors  of  the  coefficient,  and  to  represent  them  as 
multiplied  together  and  followed  by  the  several  letters, 
each  written  as  many  times  as  it  is  a  factor.  Thus, 
18  a^  m^  z=  3  .  3  .  2  «  «  m  m  w.  In  this  example,  the  dif- 
ferent prime  factors  are  3,  2,  a  and  m;  3  is  contained 
twice,  2  once,  a  twice,  and  m  three  times,  as  a  factor. 

The  same  quantity  may  be  expressed  in  its  factors  thus : 
2  .  3^  «2  Yffi^  in  which  the  exponents  show  how  many  times 
each  quantity  enters  as  a  factor. 

When  a  quantity  is  the  product  of  a  monomial  and  a 
prime  polynomial,  in  order  to  separate  it  into  factors,  it  is 
only  necessary  to  divide  it  by  the  greatest  monomial  that 
will  exactly  divide  all  the  terms,  and  to  place  the  divisor, 
separated  into  prime  factors,  before  the  quotient,  the  latter 
being  included  in  a  parenthesis. 

Thus,  am~\-  an^=^  a{m'\-n),  in  which  the  factors  are 
a  and  m  +  w.  In  like  manner,  50  a^  b^  -j"  ^^  «  b^  =  5^a  b^ 
X  (2  a  4"  ^) J  the  factors  of  which  are  5,  a,  6,  and  2  a  -f-  ^* 

Let  the  learner  separate  the  following  quantities  into 
prime  factors. 

2.  40x2y2, 

3.  120my.       ^^  -^   J     ^  -^.^^ 

4.  I5a:2y3.     J     ^  / 

5.  225 my,        '.  i     #>      ,  _ 

7.   3ax  +  7ay.       Ans.  a(Sx-(-7y). 


9iB  .  FACTORS.  [4>  16. 

f 

9.  25m3  — 577i2?^ 

10  54  a3  62  c  _  27  a  62  c^    ,  ^  ,-;^  ,:^  ^  .0^^ 

11.  81m2a:y  +  27  7w2py  — 5m3y.  ^ 

12.  44a6c  — 88a2  6  +  22a3a:.  -^a^-^j} 

13.  3m2a:2  +  6m23/2_3;/i2  ^^^y^/        ^ 

14.  30a3-|.25a26_|_5a2  -  ^^ 

Art.  60.  When  a  quantity  is  the  product  of  several 
polynomials,  the  process  of  finding  its  factors  becomes 
more  difficult ;  but  in  many  cases  some  of  the  factors  may 
be  easily  ascertained 

1.  Any  'power  of  a  polynomial  may  evidently  be  sepa- 
rated into  as  many  factors,  each  equal  to  that  polynomial, 
as  there  are  units  in  the  exponent  of  the  power.  Thus, 
x2  +  2a:3/  +  y2— (2:_|_y)2— (3;_j_^)  (x-\-y)\  and  x^ -\- 

2.  The  difference  between  the  second  powers  of  two 
quantities  can  be  separated  into  two  factors,  one  of  which 
IS  the  sum  and  the  other  the  difference  of  those  quanti- 
ties.     Thus,    T^ y'^  ^i^  {%  J^  y")  {x y)  ;    also,    X^ y^  ^= 

3.  The  difference  between  similar  powers  of  two  quan- 
tities can  be  separated  into  at  least  two  factors,  one  of 
which  is  the  difference  of  those  quantities.  Thus,  x  —  y, 
x^  —  y2^  x^  —  y3.  &,c.,  are  each  divisible  by  x  —  y. 

4.  The  difference  between  similar  even  powers  of  two 
quantities,  the  powers  being  above  the  second,  can  always 
be  separated  into  at  least  three  factors,  one  of  which  is 
the  sum,  and  another  the  difference,  of  the  quantities. 
Th\x^,m^  —  n'^—(m^-\-n^){m^  —  ri^)z=z{m^-\.n^){m-^n) 
(m  —  n). 

5.  The  sum  of  similar  odd  powers  of  two  quantities 
can  be  separated  into  two  factors,  one  of  which  is  the 


^17.]  SIMPLIFICATION    OF    FRACTIONS.  93 

sum  of  the  quantities.     Thus,   'x^^ -\- y^  =^  {x -^  y)  {x^  — 
2;  y  +  y2). 

The  quantity  x^  —  y^  can  be  separated  into  four  factors. 
Thus,  x^  —  3/^  =  {x^  +  y^)  {x^  —  y^)  =  (3^  +  y )  (x^  —  xy 

+y')(^-y)(^^  +  ^y  +  y^). 

Let  the  learner  separate  the  following  quantities  into 
prime  factors. 

1.  6«2_662.  Ans.  2.3(a  +  6)(a  — 6). 

2.  3x2  — 6xy  +  3y2.     Ans.  3(a:  — 3/)  (x  — y). 


3.    a;2_^2a;+l. 


^ 


7.    15(x^  — y6).  I 

a   25m8  — 25y8. 
9.    12m5+12/i5. 

10.  6x3— 18x2y-[-18a;y2_6y3, 

Remark.     Any  power,  also  any  root,  of  1  is  1. 

11.  9;7^7  +  9. 


SECTION   XVII. 

SIMPLIFICATION    OF    FRACTIONS. 


Art.  61.  Both  numerator  and  denominator  of  a 
fraction  may  be  multiplied  by  the  same  quantity,  without 
changing  the  value  of  the  fraction ;  for  multiplying  the 
numerator  multiplies  the  fraction,  and  multiplying  the 
denominator  divides  the  fraction ;  but  when  a  quantity  is 
multiplied,  and  the  result  is  divided  by  the  multiplier,  the 
value  of  that  quantity  remains  unchanged. 


94  SIMPLIFICATION    OF    FRACTIONS.  \^  17. 

Also,  both  numerator  and  denominator  of  a  fraction 
may  be  divided  by  the  same  quantity,  without  changing 
the  value  of  the  fraction;  for,  dividing  the  numerator 
divides  the  fraction,  and  dividing  the  denominator  multi- 
plies the  fraction  ;  but  if  a  quantity  is  divided,  and  the 
result  is  multiplied  by  the  divisor,  the  value  of  that  quan- 
tity remains  unchanged. 

Art.  63.  From  the  principle  last  stated,  we  deduce 
the  following 

RULE    FOR    SIMPLIFYING    A    FRACTION. 

Divide  the  numerator  and  denominator  hy  all  the  factors 
common  to  both. 


Simplify  the  following  fractions. 


1. 

\babc 
25  a2  62* 

Ans. 

3c 
bah 

8. 

2. 

9. 

3. 

I3a4  63 
39ax5 

10. 

4. 

45mar2i/3 
85  m3  a:  1/2' 

11. 

5. 

9a6c 
57  a2  m^' 

12. 

6, 

155  axy 

13. 

450  X 1/2 
900x3^' 
33fl6c2cP 
99a2  62c2d 
3  g  6  -f  9  tt^ 

15  fl2  52^45  0? 

9  a2  m3 
lx^y^-\-Umx^y 

21  m  x2 1/2 
m2 n -|-  6  win 


7. 


465  a2  a;  1/4  3  m^  +  w^ 

21a 


147  a2  62 

In  the  following  examples,  some  of  the  common  factors 
are  polynomials;  but  they  can  be  easily  discovered  by 
separating  the  numerators  and  denominators  into  factors. 

--      3x2  — 3 1/2  .  3  (a;  .j.  2^)  (x  —  v) 

14. -.    This  IS  the  same  as    ^    "^  ^~ -. 

6x^6y  3.2(x  — 1/) 


•J^  17. J  SIMPLIFICATION    OF    FRACTIONS.  95 

Suppressing  now  the  common  factors  3,  and  x  —  y,  we 
have  for  the  result  ^ — ^. 

2 
^^       aS^-i/3  jQ      35a2  63(a;-y) 

16      ^^-^^  20  ^^-'-'^y' 

17      9  {x^  —  y^)  c>l      5  0^4- 10aa?  + 5x8 

27(a;H-t/)*  ^  *  8a3  +  8a2x 

30a;2  +  30xi/* 

Art.  63.     When  the  division  of  one  integral  quantity 
by  another  cannot  be  exactly  performed,  it  is  expressed  in 
the  form  of  a  fraction,  the  divisor  being  placed  under  the 
dividend.     The  fraction  should  then  be  simplified. 
1.    Divide  3  a  62  c  by  6bc^. 


jCiXpressmg  me  aivision,  we 

nave ,  wnicn  reauce 

6  6  c2 

becomes  — ,  Ans. 

2  c 

2.    Divide  12  a  m^ 

by  15  a^m\ 

3.    Divide  13x4y 

by  39x2y3. 

4.    Divide  22  mx^ 

by  33m2a;. 

5.   Divide  45  a  6x3 

by  50a^b^x, 

6.   Divide  54  b^c^x 

by  33  65ca;3. 

'^,   Divide  3a6c  +  3«6m 

by  15  a^b^x. 

8.   Divide  5x2  +  5y2 

by  20(2:4  — y4). 

9.    Divide  3  (a +  6) 

by  15(a2  +  2a6  +  62), 

10.   Divide  6a^{x  +  ^) 

by  14a3(x3-|-y3). 

r  *? 


3 


96  MULTIPLICATION    OF    FRACTIONS.  [^  18 


SECTION  XVIII. 


MULTIPLICATION    OF    FRACTIONS    BY    FRACTIONS. 

Art.  64r.     Find  f  of  |^ ;  that  is,  multiply  ^  by  f . 

According  to  the  rule  for  the  division  of  fractions  by 
integers,  -^  of  -J  is  -^jj,  and,  according  to  the  rule  for  the 
multiplication  of  fractions  by  integers,  f  of  |^  is  f^,   Ans. 

Find  the  —  of  — ;  that  is,  find  the  product  of  —  by  — . 
h  d  b  d 

1  ^     n    c    .      c  J    a      ^    c    .     a  c      . 

—  part  of  —  IS  — ,  and  —  of  —  is  — ,  Ans. 
6*^  d        bd'  b  d        bd' 

T     1-,  a       m       X         amx 

in  like  manner,  —  .  —  .  —  ■=. . 

b        n       y  bny 

Hence  we  have  the  following 

RULE  FOR  MULTIPLYING  FRACTIONS  BY  FRACTIONS. 

Multiply  all  the  numerators  together  for  a  new  numer- 
ator, and  all  the  denominators  together  for  a  new  denomi- 
nator. 

Remark,  As  the  resulting  fractions  should  be  simpli- 
fied, it  is  best  to  represent  the  operation,  then  strike  out 
the  common  factors,  previous  to  the  actual  performance 

,  .    ,.        .  mi  2am    6{a-\-b)        2.6amla-\-b) 

of  the  multiplication.    Thus,  .  = »^         ' 

a{a-{-b) 

mxy 

1.  Multiply  by  . 

^  •'    4>xy  -^     9m 

2.  Multiply  by 


4, by*  "^    25 o3 jc 

3.    Multiply  -^  by  — . 


<^  19.]         LEAST  COMMON  MULTIPLE.  97 

4.    Multiply  — -^--  by . 

6.    Multiply  — ^--^  by  . 

6.    Multiply  ?^::^  by     '"^ 


4  62  -^     x-\-y 

7.  Multiply  ^-^=^"  by  ii<^-^. 

8.  Find  the  product  of  — , ,  and  -^. 

^  6  '  4cd'  7m2 

9.  Find  the  product  of  '-i^^,  '-^^^,  and  — ^— . 
10.    Find  the  product  of  — —,  ^  "~   ^  ,  and 


146c'  13  '  4(a-6)* 


)^ 


SECTION   XIX, 


LEAST    COMMON    MULTIPLE 


Art.  OS*  When  one  quantity  is  divisible  by  another, 
the  former  is  called  a  multiple  of  the  latter.  Thus,  10  is 
a  multiple  of  5 ;  it  is  also  a  multiple  of  2. 

A  common  multiple  of  two  or  more  quantities  is  one 
which  is  divisible  by  tiiem  all ;  and  the  least  common 
multiple  is  the  least  quantity  divisible  by  them  all.  Thus, 
24  is  a  common  multiple  of  6.  and  4,  but  12  is  the  least 
common  multiple  of  these  numbers. 

Let  it  be  required  to  find  the  least  common  multiple  of 
Sa^m  2Lnd*6am^. 

It  is  manifest  that  this  multiple  must  contain  all  the 
factors  of  Sa^m  and  6am^.     Separating  these  quantities 


98  LEAST    COMMON    MULTIPLE.  [^1^ 

into  their  prime  factors,  we  have  8  a^  m  zzz  2^  a^  ^ji^  ^nd 
6  a  m^  ::;p  2  .  3  a  m^.  The  different  prime  factors  are  2,  3, 
a,  and  wi,  each  of  which  must  be  contained  in  the  multi- 
ple required,  as  many  times  as  it  is  found  in  either  of  the 
given  quantities ;  that  is,  2  must  be  contained  three  times, 
3  once,  a  twice,  and  m  three  times,  as  a  factor.  The 
least  common  multiple  is,  therefore,  2^.  3  a^ m^^  or  24  a^ m^ 

Art.  66.     Hence  we  have  the  following 

RULE     FOR     FINDING    THE     LEAST     COMMON    MULTIPLE     OP 
SEVERAL    QUANTITIES. 

First,  separate  the  quantities  into  their  prime  factors ; 
then  unite  in  one  product  all  these  different  factors,  each 
raised  to  the  highest  power  found  in  either  of  the  given 
quantities. 

Find  the  least  common  multiple  in  each  of  the  follow- 
ing examples. 

1.  4a^l0«63.  Ans.  22.5a263  — 20a263 

2.  6m^x,  8  m  2/^, 

3.  4a:3y,  2xy^,  9m^, 

4.  25,  15m2,45x3m. 

5.  Sxt/,  15x3y2^3(a  +  6). 

6.  llp^q^,S3py,22pq^, 

7.  7(a-f6),  14(«2  +  2a6  +  62) 

8.  2a6  +  46^  and  Ua^b^^ 
9    18(x  — y),9(x2— y2). 


^  20.]    ADDITION  AND  SUBTRACTION  OF  FRACTIONS.      99 


SECTION   XX. 

ADDITION     AND     SUBTRACTION     OF     FRACTIONS.         COMMON 
DENOMINATOR. 

Art.  67.     The  addition  and  subtraction  of  fractions 
are  represented  by  writing  them  after  each  other  with  the 

sign  +  and  —  between  them,  thus,  —-\ — ,  care 

being  taken  to  place  the  signs  even  with  the  line  which 

separates  the  numerator  from  the  denominator. 

But  when  the  denominators  are  alike,  we  may  perform 

the  addition  or  subtraction  upon  the  numerators,  placing 

2  3 

the  result  over  the  common  denominator.     Thus, [-  — 

2+3         5  1     «     I      6  a+b        1  5  2         5—2 

=1 =z  — ,  and [--—  = ;  also, = 

7  7'  m     *     m  wi'  '9  9  9 

3  1  J     "  ^  "  —  ^ 

9  '  3  '  m  m  m     ' 

Suppose  it  were  required  to  add  —  and  — .     In  this 

case  the  denominators  are  different ;  but  if  the  numerator 
and  denominator  of  the  first  fraction  be  multiplied  by  n, 
and  the  numerator  and  denominator  of  the  second  fraction 
be  multiplied  by  m,  the  denominators  of  the  fractions  will 
be  made  alike,  while  the  value  of  the  fractions  remain  un- 
changed.    The  first  fraction  then  becomes  —  ,  and  the 

o  win ' 

second  becomes   — ,  by  the  addition  of  which  we  have 
Also,  the  difference  of  the  same  fractions  is 


an-\'  bm 


m  n 


100    ADDITION  AND  SUBTRACTION  OF  FRACTIONS.   [§  20 

Let  It  be  required  to  add  — ,  — ,  and  — .     First,  we  re- 
ft     m  4/ 

duce  the   fractions  to   a  common   denominator.     If  the 

numerator  and  denominator  of  each  fraction  be  multiplied 

bv  the  denominators  of  both  the  others,  which  does  not 

change  the  value  of  the  fractions,  they  become  — -,  — -, 

^  bmy    bmy 

,    bmx      ,  /»      1  •    1    •      amy4-bcy4-bmx      . 

and  ,  the  sum  of  which  is ,  Ans. 

6  my  bm y 

Hence  we  derive  a 

RULE  FOR  THE  ADDITION  AND  SUBTRACTION  OF  FRACTIONS. 

Reduce  them  to  a  common  denominator,  then  add  the 
numerators,  or  subtract  one  from  the  other,  placing  the 
result  over  the  common  denominator. 

Art.  68.  From  the  preceding  examples,  we  derive 
also  the  following 

RULE     FOR    REDUCING    FRACTIONS    TO    A    COMMON 
DENOMINATOR. 

Multiple/  all  the  denominators  together  for  a  common 
ienominator,  and  multiply  each  numerator  by  all  the  de- 
nominators except  its  own,  in  order  to  obtain  the  numer- 
ators. 

The  results  obtained  by  this  rule  are  correct,  but  not 
always  the  simplest. 

Let  us   reduce   — ,  — —,  and   to  a  common  de- 

nominator.  By  taking  the  product  of  all  the  denomina- 
tors, we  shall  obtain  a  common  denominator  considerably 
greater  than  is  necessary.  In  this  case,  the  least  common 
denominator  will  be,  as  in  arithmetic,  the  least  common 
mu  tiple  of  the  given  denominators.     The  least  common 


<^  20.]  COMMON    DENOMINATOR.  ^01 

multiple  of  3  a^,  4a 6,  and  6b^c  is  3 . 2^  a^ 6^ c,  or  12  a^ 6^ c, 
which  is  the  least  common  denominator  sought.  To  pro- 
duce 12  a^  b^  c,  the  first  denominator  is  multiplied  by  4  6^  c, 
the  second  by  3  a  6^  c,  and  the  third  by  2  a^ ;  these  are, 
therefore,  the  quantities  by  which  the  respective  numera- 
tors are  to  be  multiplied,  and  are  evidently  obtained  by 
dividing  the  common  denominator  by  each  of  the  given 
denominators  separately.  This  multiplication  being  per- 
formed, and  the  results  placed  over  the  common  denomi- 

,       ^        .          ,                   4  63  c  a?      9ab^cy          ,     lOa^w 
nator,  the  fractions  become    ,  -,  and  . 

'  12a263c'   12a263c  I2a^b^c 

Art.  09.     Hence  we  have  the  following 

RULE     TO     REDUCE      FRACTIONS     TO     THE     LEAST     COMMON 
DENOMINATOR. 

JFHnd  the  least  common  multiple  of  all  the  given  denomi" 
nators,  and  this  will  be  the  least  common  denominator; 
then  divide  the  common  denominator  by  each  of  the  given 
denominators,  and  multiple/  the  numerators  by  the  re-- 
spective  quotients,  placing  each  of  these  products  over  the 
common  denominator. 

Remark.  Fractions  must  be  simplified  before  applying 
this  rule. 

1     Reduce  — ,  — ,  and  —  to  a  common  denominator. 
h     d  y 

.  a  dy    b  c  y  ,  b  dx 

Ans.   — -,  — -,  and  . 

b  dy    b  dy  b  dy 

2.  Reduce  — , ,  and  — -  to  the  least  common  de- 

lab    14  a2  2163 

nominator.  Ans.  — r — , ,  and  -. 

42  a2  63'  42  a2  6^'  4S  a2  ^3 

3.  Reduce  -—  ,  and  — ^  to  the  least  common  denomi- 

7  wi3  21  a  6 

nator. 


\^\ 


102    ADDITION  AND  SUBTRACTION  OF  FRACTIONS.    [<§>  20* 

^   4.    Reduce  —  — ,  -^,  and  —  to  the  least  com- 

6m2x     9m3  21  wi  a:2 

men  denominator. 

\    5.    Add  —  and  — . 

2 1/  6c 

6.  Add  —   and   . 

3  m2  9  w  x* 

7.  Add  and 


75  a;2  25  xS  i/^ 

.8.    Add  ^  and  -'^  , 
9.   Add  ^=^  and  ^±-*. 

27  81 


y 


10. 

Add and  . 

1  +  a             i_a 

11. 

Add  and  . 

a;2  —  2/2             a;  „  2/ 

12. 

From  -^  subtract  -^. 

3                            8 

13. 

From  '"^  subtract''^. 
la                       lOa* 

14. 

From  subtract  

1— x  1  +  a; 

15.  From  —  subtract  — ^^^. 

la  2\a^ 

16.  From  subtract  -—^. 

a^  —  b^  a-\-b 

-^T^  3a;2  ,  76c 

17.  rrom  subtract 


4  w^ n  wi2  -f  win 

18.    Reduce  2|,  that  is,  2  +  f ,  to  the  form  of  a  fraction 
Since  4  fourths  make  1,  2  =  f ;  and  f  -j-  f  iz=  ^^  Ans 
In  like  manner,  7  —  f  :=  -^^6-  —  |  z=  -&^. 


19.    Add  b  and    -. 


Reducing  6  to  a  fraction  having  d  for  a  denommator 

,  I        b  d  ,    b  d    .     c         b  d  4-  c      . 

we  have  6  =  —  ;  and  —  H = ,  Ans. 

rf  '  d    ^    d  d     ' 


^21.]  DIVISION    BY    FRACTIONS.  108 

20.  Reduce  3  a  +  —  to  a  fraction. 

3  X 

21.  Reduce  1  nv^ — • —  to  a  fraction. 

41/ 

22.  Reduce  3  X ^  to  a  fraction.    Ans  — ~^ 

7  " 

23.  Add  a4-b  and  ^^^. 

y  4 

24.  From  ^^-^^ —  subtracts  —  a, 

9  a?  I  4  1/ 

25.  Reduce  — -^-— ^  +  9  x  y  to  a  fraction. 

26.  Reduce  12^2 iH — ^  to  a  fraction. 

3a 

27.  From  ^  "^  ^    subtracts  —  1. 

28.  Reduce    ^   "*"    ^  +  2  c  to  a  fraction. 
4r  ' 


SECTION   XXI. 

DIVISION    OF    INTEGRAL    AND    FRACTIONAL    QUANTITIES    RV 
FRACTIONS. 

Art.  70.     How  many  times  io  f  contained  in  9? 

Since  in  9  there  are  ^f^,  |  is  contained  63  times  in  9, 
and  ^  is  contained  ^  as  many  times,  that  is,  ^-^;  in  other 
word**,  9  divided  by  ^  gives  ^-^-  for  a  quotient.  The  result 
is  the  .same  as  the  product  of  9  by  ^. 

How  many  times  is  f  contained  in  a  ? 

Since  in  a  number  a  of  units  there  are  5  a  fifths,  ^  is 
contained  5  a  times  in  a,  and  f  is  contained  ^  as  many 


DIVISION    BY    FRACTIONS.  [^  ^1» 

5  a 
times,  that  is,  —  times ;  in  other  words,  a  divided  by  | 

4 

gives  —  for  a  quotient.     This  result  is  the  same  as  the 

product  of  a  by  f . 

Hovv^  many  times  is  —  contained  in  a  ? 

n 

Since  in  a  number  a  of  units  there  are  — ,  —  is  con- 
tained  na  times  in  a,  and  —  is  contained  —  as  many 
times,  that  is,  —  times ;  or  a  divided  by  —  gives  —  for 

a  quotient.     This  is  the  same  as  a  .  — . 

How  many  times  is  f  contained  in  y\  ? 

Reducing  the  fractions  to  a  common  denominator,  we 
have  f  =  f f ,  and  -^^  =  ff-  ^^^  f f  is  contained  in  f f  as 
many  times  as  44  is  contained  in  63,  which  is  f  f ;  or  ^^ 
divided  by  f  gives  ff  for  a  quotient.  This  result  is  the 
same  as  y\  .  f. 

How  many  times  is  —  contained  in  — ? 

Reducing  the  fractions  to  a  common  denominator,  we 

,  a         a  71  ,wi         6771       -rj^an.  .       ^  .      bm 

have  —  =  — ,  and  —  =  — .     J5ut  —  is  contamed  m  — 

b         b  n  n  bn  b  n  b  n 

as  many  times  as  a  /i  is  contained  in  b  m,  that  is,  —  times  ; 
or  —  divided  by  —  gives  —  for  a  quotient.  Tie  result 
IS  the  same  as  —  .  — . 

n        a 

From  the  solution  of  the  preceding  questions  we  derive 
the  following 

RULE    FOR    DIVIDING    ANY    QUANTITY    BY    A    FRACTION. 

Ivvert  the  divisor ,  and  then  proceed  as  in  multiplication 


<§>21.]  DIVISION    BY    FRACTIONS.  105 

Remark,  When  we  wish  to  find  what  part  one  quantity 
is  of  another,  we  make  the  quantity  called  the  part  the 
dividend,  and  the  other  the  divisor ;  also,  when  we  wish 
to  find  the  ratio  of  one  quantity  to  another,  we  make  the 
quantity  mentioned  first  the  dividend,  and  the  other  the 
diviso" 

Perforin  the  following  questions,  recollecting  to  sim- 
plify, as  directed  in  Article  62.  Thus,  dividing  —  by 
6am  ,  4.56cm        2.56c        10  6c 

,  we  have  — = =: . 

56c  G.lamn.       S.Ian        '21  an 


1. 

Divide  a 

by 

4 

2. 

Divide  my 

by 

2a 
6* 

3. 

Divide  ^m^y'^ 

by 

4>my 
3x  ' 

4. 

Divide  5a2_ 

4  62 

by 

3mn 
7a;2 

6. 

Divide  3  (a +  6) 

by 

6a2 
bxy 

6. 

Divide  4,ahc^ 

by 

8a2 
9  m  a;  +  3  v' 

7. 

Divide  lla^^ 

-22x2 

by 

33  a; 
4i/ 

8. 

Divide  — 
y 

by 

3 

7' 

9. 

Divide  — 

4t/2 

by 

6a;2 
5a6* 

10 

Divide''"*"^ 

xy 

by 

3x 

/ii. 

Divide  *^"^^ 

3m 

by 

2a;y 
9m2' 

12. 

^.    .,      5a2a;2 
Divide  

by 

lOax 

9  m3  2/3  •'    27  m  ^ 


106  EQUATIONS    OF    THE    FIRST    DEGREE  [^S*  22 

13.    m  is  what  part  of  —1 


m 


14.    is  what  pa.t  of  —1 

m-\-n  y 

15.  ^±i!- is  what  part  of-? 

8ab  In 

16     What  is  the  ratio  of  ^^^±^  to  ^^^-^  ? 
17.    What  is  the  ratio  of  ?i^^  to  ^i^^^? 

w  +  n  7  0:3 

18     What  IS  the  ratio  of  — ^^ to  — r^- — -1 

l{x  +  y)  21(a;2  — r«^ 


SECTION   XXII. 

EQUATIONS    OF    THE    FIRST    DEGREE    CONTAINING    TWO      \  \  /'\^ 
UNKNOWN    QUANTITIES. 


Art.  71,  When  a  question  involves  several  inde- 
pendent unknown  quantities,  in  order  that  we  may  be 
able  to  determine  them,  there  must  be  given  as  many 
conditions,  and,  consequently,  we  must  be  able  to  form 
as  many  independent  equations,  as  there  are  unknown 
quantities. 

1.  A  man  gave  27  sT  for  3  bushels  of  corn  and  4 
bushels  of  oats  ;  and,  at  the  same  rate,  he  gave  50  s.  for 
7  bushels  of  corn  and  5  bushels  of  oats.  What  was  the 
price  of  each  per  bushel  ? 

Let  X  (shill.)  =  the  price  of  corn,  and  y  (shill.)  =r  the 
orice  of  oats,  per  bushel.     Then, 


^22.]     CONTAINING    TWO    UNKNOWN    QUANTITIES.        107 

{1)    3  a:  +  4  y  =  27 ;  and  )  Multiply  the  1st  by  7,  and  the 

(2)  72:-f5y  =  50.  /  2d  by  3, 

(3)  21a:+28yz=189;    ) 

(4)  21  a:  +  15y  =:  150.     I  Subtract  the  4th  from  the  3d, 
21a:  +  28y  — 21a:— 15y=  189—150;  reduce, 

133^  — 39.-.yi=3s. 
Substitute  3  for  7/  in  the  1st, 
3  2:  +  12  =  27 ;  .-.  3  2:  =1 27  —  12, .-.  3  a:  z=  15  .-.  a:  zz:  5  s. 
Ans.     Corn,  5  s.,  rye,  3  s.  per  bushel. 
We  might  have  multiplied  the  1st  equation  by  5,  and 
the  2d   by  4,  and  then  have  subtracted  one  of  the  re- 
sulting equations  from  the  other.     We  should  thus  have 
obtained  an  equation  without  1/,  and  from  this  we  could 
have  found  the  value  of  x.     This  value  of  a;,  substituted  in 
one  of  the  first  two  equations,  would  have  given  the  value 
of  y. 

2.  The  sum  of  two  numbers  is  20,  and  if  twice  the 
'ess  be  subtracted  from  3  times  the  greater,  the  remainder 
will  be  25.     Required  the  numbers. 

Let  X  z=  the  greater,  and  y  =  the  less.     Then, 

(1)  a:+    y=:20;^ 

(2)  3  a:  — 2y  zz=25.  /  Multiply  the  1st  by  3, 

(3)  3  X  +  3  y  nz  60  ;  subtract  the  2d  from  the  3d, 
3a:4-3y  — 3a:  +  2y=:60  — 25;  reduce, 

5y  =  35.-.y=:7. 

Substitute  7  for  y  in  the  1st. 

a: +  7  =  20,  .•.a:=13. 

Ans,     13  the  greater,  and  7  the  less. 

This  question  may  be  performed  in   another  way,  as 

follows. 

The  first  two  equations  are 

(1)  a:+    y  =  20;) 

(2)  3a:  — 2y  =  25.  )  Multiply  the  1st  by  2, 


lOS  EQUATIONS    OF    THE    FIRST    DEGREE  [§  22. 

(3)   2 x  + 2^  =  40;  add  the  2d  and  3d, 
3x  — 2y  +  22;  +  2y==25  +  40;  reduce, 

Substituting  13  for  x  in  the  1st, 

13  +  yzz:20,  .-.^=17. 

3.  A  farmer  sold  6  barrels  of  apples  and  8  barrels  of 
pears  for  $  22 ;  also,  at  the  same  rate,  9  barrels  of  apples 
and  24  of  pears  for  $  57.  Required  the  price  of  each  per 
barrel. 

Let  X  (dollars)  1=  the  price  of  apples,  and  y  (dollars)  =  the 
price  of  pears  per  barrel.     Then, 

(1)  6.;+    8yz=22;^ 

(2)  9  X  +  24  y  zz:  57.  )  Multiply  the  1st  by  3, 

(3)  18  X  +  24  y  =:  66 ;  subtract  the  2d  from  the  3d, 
9  a:  =  9,  .*.  a:  =  $1,  price  of  apples  per  barrel. 

Substitute  1  for  x  in  the  1st, 
6  +  8  y  =1 22,  .-.  y  =  $2,  price  of  pears  per  barrel. 

Since  18  is  the  least  common  multiple  of  6  and  9,  if 
we  had  multiplied  the  1st  by  3,  and  the  2d  by  2,  we 
should  have  obtained  two  equations  in  which  the  coeffi- 
cients of  X  would  have  been  alike.  Then,  after  sub- 
traction, we  should  have  found  the  value  of  y,  which,  by 
substitution,  would  have  given  that  of  x. 

But  we  will  solve  this  problem  in  another  way,  which  is 
sometimes  a  convenient  one.  Resuming  the  first  two 
equations, 

(1)  6x-f   8y=i22;  )  Divide  the  1st  by  2,  and  the  2d 

(2)  9x  +  24yz=57.  )  by  3, 

(3)  3x+    4yz=ll;) 

(4)  3  X  +    8  y  =  19.  /  Subtract  the  3d  from  the  4th, 

4  y  =  8  .'.  y  =  $2,  price  of  pears  per  barrel. 
Substituting  2  for  y  in  the  3d, 
Sx-^S^ill  r.x=^  $1,  price  of  apples  per  barrel 


<§>22.]      CONTAINING    TWO    UNKNOWN    QUANTITIES.       109 

4.  Two  numbers  are  such,  that  if  J  of  the  first  be  in* 
creased  by  10,  the  sum  will  be  equal  to  §  of  the  second ; 
and,  if  f  of  the  second  be  diminished  by  10,  the  remain- 
der will  be  equal  to  -^q  of  the  first.  Required  the  num- 
bers. 

Let  X  =  the  1st,  and  y  :=  the  2d.     Then, 

(1)   I  +    10  =  ^ 

4,y  ix    ^Clearing  the  equations  of  frac- 

^  ^     6  10     y  tions, 

(3)  3x+   60==4y;) 

(4)  8y  — 100z=:7a;.    i 

In  the  3d  and  4th,  transpose  the  unknown  terms  into 
the  1st  members,  and  the  known  into  the  2d, 

(5)  Sx  —  4:7/  =  —   60;  ) 

(6)  Sy  —  7xz=z       100.  )  Multiply  the  5th  by  2, 

(7)  6x  —  Sy=i  — 120.     Add  the  6th  and  7th, 

8y  —  7a:  +  6a;  — 8^=100  —  120;  reduce, 
—  x  =  — 20  .•.  changing  the  signs,  a:  =  20,  the  1st  num- 
ber. 
Substitute  20  for  x  in  the  5th, 
60  — 4^^  =  — 60,  or  4y  — 60  =  60  .•.3/  =  30,   the  2d 
number. 
We  perceive  that,  in  the  preceding  problems,  the  con- 
ditions of  each  gave  rise  to  two  independent  equations, 
which   may  be  called  the  original  equations.     The  suc- 
ceeding  equations    were    deduced    from,    or   were    mere 
modifications  of,  the  original  equations. 

From  the  two  original  equations  containing  two  un- 
known quantities,  one  was  obtained  containing  only  one 
unknown  quantity.  The  process  by  which  this  is  done  is 
called  elimination.  It  is  eliminating  that  unknown  quan- 
tity which  the  new  equation  does  not  contain.  For  ex- 
ample, in  the  1st  question  of  this  Article,  we  eliminated  x. 


110  EQ,UATIONS    OF    THE    FIRST    DEGREE  [^22 

and  obtained  an  equation  containing  no  unknown  quantity 
except  y. 

We  perceive,  from  what  precedes,  that,  when  the  un- 
known quantity  to  be  eliminated  is  found  in  corresponding 
members  in  both  equations,  that  is,  in  the  first  members  of 
both,  or  in  the  second  members  of  both,  the  elimination 
can  be  effected  according  to  the  following  rule. 

FIRST    METHOD    OF    ELIMINATION. 
RU  L  E  . 

Multiply  or  divide  the  equations,  if  necessary,  so  as  to 
make  the  coefficients  of  the  quantity  to  be  eliminated  alike 
in  the  two  equations ;  then  subtract  one  of  the  resulting 
equations  from  the  other,  if  the  signs  of  the  terms  con^ 
taining  this  quantity  are  alike  in  both  equations ;  but  add 
them  together  J  if  the  signs  are  different. 

Previously  to  applying  this  rule,  it  is  advisable  to  free 
the  equations  of  fractions,  if  they  contain  any ;  and  to 
transpose  all  the  unknown  terms  into  the  first  members, 
and  the  known  terms  into  the  second.  Moreover,  if,  in 
either  equation,  there  are  several  terms  containing  the 
unknown  quantity  to  be  eliminated,  these  terms  should  all 
be  reduced  to  one. 

The  coefficients  of  any  letter  in  the  two  equations  will 
be  made  alike,  if,  after  the  equations  are  prepared  as  pre- 
scribed above,  each  equation  be  multiplied  by  the  coeffi- 
cient of  that  letter  in  the  other ;  or,  if  each  equation  be 
multiplied  by  the  number  by  which  the  coefficient  of  that 
letter  in  this  equation  must  be  multiplied,  in  order  to  pro- 
duce the  least  common  multiple  of  the  two  coefficients  of 
the  letter  to  be  eliminated. 

Thus,  in  the  3d  example,  the  least  common  multiple  of 
6  and  9,  the  coefficients  of  x  in  the  1st  and  2d  equations, 
is  18,  which  may  be  produced  by  multiplying  6  by  3,  or  9 


<§*  22.]      CONTAINING    TWO    UNKNOWN    Q,UANTITIES.       Ill 

by  2.  If,  then,  the  1st  equation  be  multiplied  by  3,  and 
the  2d  by  2,  the  coefficients  of  x  will  be  alike  in  the  two 
resulting  equations. 

5.  Four  bushels  of  wheat  and  3  bushels  of  corn  cost 
$11 ;  and,  at  the  same  rate,  5  bushels  of  wheat  and  6  of 
corn  cost  $16.     Required  the  price  of  a  bushel  of  each. 

6.  Twice  A's  money  and  5  times  B's  make  $165;  also 
6  times  A's  money  and  7  times  B's  make  $295.  How 
much  money  has  each  ? 

7.  A'draper  bought  2  pieces  of  cloth  for  £St  15  s.,  one 
at  4  s.  per  yard,  and  the  other  at  5  s.  He  sold  the  whole 
at  an  advance  of  2  s.  a  yard,  and  thereby  gained  £\  4  s. 
How  many  yards  were  there  in  each  piece  ? 

8.  Four  cows  and  1  sheep  cost  $  82 ;  also  20  sheep 
and  1  cow  cost  $60.  Required  the  cost  of  a  cow  and 
that  of  a  sheep. 

9.  A  farmer  gave  IQ  bushels  of  corn  and  $2*50  for  7 
bushels  of  wheat ;  he  also  gave  15  bushels  of  corn  for  5 
bushels  of  wheat  and  $4*50.  What  was  the  estimated 
worth  of  a  bushel  of  each  ? 

"^^  10.  The  sum  of  ^  of  A's  age  and  \  of  B's  is  40  years  ; 
and  if  ^  of  A's  age  be  subtracted  from  B's,  the  remainder 
will  be  K>K>  years.     Required  the  age  of  each. 

11.  Two  drovers,  A  and  B,  counting  their  sheep,  found 
that,  if  A  had  10  more  and  B  10  less,  their  flocks  would 
be  equal ;  but  if  B  had  10  more  and  A  10  less,  A  would 
have  only  f  as  many  as  B.     How  many  sheep  had  each  ? 

12.  A  man  wrought  o  days,  havmg  his  son  with  him  4 
days,  and  received  for  their  joint  labor  $8;  he  afterwards 
wrought  8  days,  having  his  son  with  him  6  days,  and  re- 
ceived $11.     llequired  the  daily  wages  of  each. 

Art.  73.  1  The  sum  of  two  numbers  is  13,  and 
their  diiference  is  3.     Required  these  numbers 


112  EQUATIONS    OF    THE    FIRST    DEGREE  [*§>  22. 

Let  X  =:  the  greater,  and  y  =  the  less.     Then, 

(1)  x  +  y=13,) 

(2)  X  —  y  =  3.     i  Transpose  y  in  both  1st  and  2d, 

(3)  x=13  —  y; 

(4)  z  =  3-\-j,. 

Since  the  2d  members  of  the  3d  and  4th  are  each  equal 
to  z,  they  are  equal  to  each  other  (Ax.  7) ;  .•. 
3  -f-  y  zz:  1 3  —  y ;  .•.  2  y  =  10,  and  y  =  5,  the  less  number. 
Substitute  5  for  y  in  the  4th, 
a;  zz:  3  -j-  5  =  8,  the  greater  number. 

2.  If  3  yards  of  linen  and  4  yards  of  cotton  cost 
$2*85,  and,  at  the  same  rate,  5  yards  of  linen  and  7 
yards  of  cotton  cost  $4*80,  what  is  the  price  of  each  per 
yard? 

Let  X  (cents)  zz:  the  price  of  linen,  and  y  (cents)  zz:  that  of 
cotton,  per  yard.     Then, 

(1)  3a:4-4y  zi:285;  >  Transpose  4y  in  1st,  and  divide 

(2)  5x  +  7y  =  480.  i  by  3, 

285  •—  4 1/ 

(3)  x=z -.     Transpose  7y  in  2d,  and  divide  by  5. 

(4)  ,  =  ^2=11, 

5 

Put  the  values  of  x  in  3d  and  4th  equal  to  each  other, 

285— -42/        480—7  2/       nr    i  •    i     i       r»         -,  ^  -■  ^' 

=: ^.     Multiply  by  3  and  5,  or  15, 

3  6 

1425  —  20  y  z=  1440  —  21  y.     Transpose  and  reduce, 

y  z=  15  cents,  the  price  of  a  yard  of  cotton. 

Substitute  15  for  y  in  the  3d, 

X  = =.  75  cents,  price  of  a  yard  of  linen. 

3.  What  fraction  is  that,  from  the  numerator  of  which 
if  1  be  subtracted,  the  value  of  the  fraction  will  be  f ; 
but  if  4  be  added  to  the  denominator,  the  value  of  the 
fraction  will  be  ^  ? 


•JjSS.]     containing  two  unknown  quantities.     113 
Let  X  =  the  numerator,  and  y  =  the  denominatoTc    Then, 

(')  T^=H 

(^)  '^^  =  \'  )  Multiply  the  1st  by  y, 

y3)  X  —  1  nz  — ^ ;  transpose  —  1, 

(4)  a;  =  5^+1.     Multiply  the  2d  by  y  + 4, 

(5)  X  =  ' .     Put  the  values  of  x  in  4th  and  5th  equal, 

3y  _|_  J  _. ^+i      Multiply  by  10, 

6y  +  10=:5y-f~^^-     Transpose  and  reduce, 

y=:10,  the  denominator. 

Substitute  10  for  y  in  the  5th 

10  +  4       ^      , 

X  = =:  7,  the  numerator. 

The  fraction  sought,  therefore,  is  -^jj. 
The  solution  of  the  three  preceding  questions  has  been 
effected  as  follows,  viz. ;  we  first  found  the  value  of  x  from 
each  of  the  original  equations,  as  if  y  were  known ;  that 
is,  we  found  from  each  equation  an  expression  for  x,  con- 
sisting of  y,  and  known  numbers ;  then,  by  equalizing 
these  two  values  of  a;,  we  obtained  an  equation  without  x, 
from  which  we  determined  the  value  of  y.  We  might 
have  eliminated  y  in  a  similar  manner,  and  found  an 
equation  without  that  letter.     Hence  we  have  a 

second  method  of  elimination. 

RULE. 

JFHnd  the  value  of  one  of  the  unknown  quantities ,  from 
each  of  the  equations,  as  if  the  other  unknown  quantity 
were  determined ;  then  form  a  new  equation  hy  putting 
these  two  values  equal  to  each  other. 

8 


114  EQUATIONS    OF    THE    FIRST    DEGREE  [<§>  22. 

Observe,  however,  that  the  unknov^n  quantity  itself 
must  not  be  contained  in  any  expression  for  its  value. 

Let  the  learner  solve  the  follovi^ing  problems  according 
to  this  second  method. 
J  4.  A  grocer  paid  $18  for  4  barrels  of  beer  and  3  bar- 
rels  of  cider;  and,  at  the  same  rate,  he  paid  $27  for  5 
barrels  of  beer  and  6  barrels  of  cider.  Required  the  price 
of  each  per  barrel. 

5.  A  carpenter  received  for  5  days'  labor  of  himself, 
'    and  3  days'  labor  of  his  journeyman,  $14*50  ;  but  he  him- 
self earned  $8  more  in  7  days  than  his  journeyman  did 
in  4  days.     Required  the  daily  wages  of  each. 

6.  Says  A  to  B,  *^  f  of  my  money,  and  f  of  yours,  make 
$55 ;  and  \  of  my  money,  increased  by  $10,  is  equal  to  ^ 
of  yours  diminished  by  $5."    How  much  money  has  each? 

7.  If  A  gives  B  $10  of  his  money,  they  will  hav  equal 
sums ;  but  if  B  gives  A  $10  of  his  money,  he  will  then 
have  only  f  as  much  as  A.  Required  the  money  of 
each. 

8.  A  market-man  bought  eggs  at  2  for  a  cent,  also 
some  at  8  for  5  cents,  giving  for  the  who^e  50  cents,  and 
sold  them  all  at  a  cent  apiece,  gaining  on  the  whole  40 
cents.     How  many  of  each  kind  did  he  buy? 

9.  There  is  a  fraction  such  that,  if  its  numerator  be 
increased  by  1,  the  value  of  the  fraction  will  be  ^;  but  if 
the  denominator  be  increased  by  3,  the  value  of  the  frac- 
tion will  be  \,     Required  the  fraction. 

10.  I  can  buy  4  pounds  of  beef,  and  6  pounds  of  mut- 
ton, for  76  cents,  and  I  find  that  8  pounds  of  beef  are 
worth  8  cents  more  than  12  pounds  of  mutton.  Required 
the  price  of  each  per  pound. 

11.  A  grocer  mixes  tea  at  3  s.  with  tea  at  5  s.  per  lb., 
and  finds  the  whole  mixture  worth  .£3  lis.;  but  3  times 
the  number  of  lbs.  of  the  first  kind  is  1  lb.  more  than 


.t 


.0 


§.2^.]       CONTAINING    TWO    UNKNOWN    QUANTITIES.       115 

twice  the  number  of  lbs.  of  the  second  kind.     How  many 
lb?   of  each  did  the  mixture  contain  ? 

Art.  73.     1.   Two  pairs  of  boots,  and  1  pair  of  shoes, 
cost  $12 ;  and  1  pair  of  boots,  and  two  pairs  of  shoes,  cost 
$  9.     Required  the  price  of  each  per  pair. 
Let  X  (dolls.)  =L  the  price  of  a  pair  of  boots,  and  y  (dolls.)^ 
zi:  that  of  a  pair  of  shoes.     Then, 

(1)  2x  +  3/=12;) 

(2)  x  +  2y  =  9.     i  Transpose  1/  in  1st,  and  divide  by  2 

(3)  ^  =  '^. 

Now,  since  — ^^  is  the  value  of  x,  this  value  may  be 

substituted  instead  of  x  in  the  2d  equation. 
We  then  have 

3  y  =  6,  and  y  zzi  $  2,  price  of  a  pair  of  shoes. 
Substitute  2  for  y  in  3d, 

X  =: =  $  5,  price  of  a  pair  of  boots. 

2.  There  are  2  numbers,  such  that  3  times  the  less, 
and  twice  the  greater,  make  120 ;  and  if  5  times  the  less 
be  subtracted  from  4  times  the  greater,  the  remainder  will 
be  20.     Required  the  numbers. 

Let  X  zz:  the  greater,  and  y  =:  the  less.     1  hen, 
(1)   2a:  +  3y  =  120; 


y  =  120;) 

y  =  20.     i 


(2)  4a;  —  5y  =  20.     '  From  the  1st  we  have 

]  20 2  X 

(3)  3/  = .     Multiply  this  value  of  y  by  5,  and  sub- 

fion  •_  in  'r 
stitute  the  result ,  instead  of  5  y  in  the  2d.     But 

since  5  y  in  the  2d  is  subtracted,  tiiis  value  of  5y,  when 
it  is  substituted,  must  also  be  subtracted.  Making  this 
substitution,  we  have 


116  EQUATIONS    OF    THE    FIRST    DEGREE  ^"^y  22. 

4  X      — — =  20.     Clearing  this  of  fractions,  transpos- 
ing, reducing,  and  dividing, 
X 1=  30,  the  greater. 
Substituting  30  for  x  in  the  3d, 

y  = 1=  20,  the  less. 

From  the  solution  of  the  two  preceding  questions,  we 
derive  a 

THIRD    METHOD    OF    ELIMINATION. 


Findf  from  one  of  the  equations,  the  value  of  the  quan* 
tity  to  he  eliminated,  as  if  the  other  unknown  quantity 
td^ere  determined,  and  substitute  this  value  in  the  other 
equation,  instead  of  the  unknown  quantity  itself 

Let  the  following  questions  be  performed  according  to 
the  third  method  of  elimination. 

3.  A  boy  bought  3  pears,  and  2  peaches,  for  18  cents  ; 
his  companion  bought,  at  the  same  prices,  7  pears,  and  5 
peaches,  for  44  cents.  Required  the  price  of  a  peach, 
also  that  of  a  pear. 

4.  If  7  lbs.  of  butter  and  6  lbs.  of  sugar  cost  $1'53, 
and  10  lbs.  of  butter  and  12  lbs.  of  sugar  cost  $2*46, 
what  is  the  price  of  each  per  lb.  ? 

5.  A  banker,  having  two  drawers  containing  money, 
found  that  if  he  transferred  $100  from  the  first  drawer  to 
the  second,  the  former  would  contain  f  as  much  money 
as  the  latter ;  but  if  he  transferred  only  $50  from  the  first 
to  the  second,  they  would  then  contain  equal  sums.  How 
much  money  was  there  in  each  ? 

6.  Two  boys,  talking  of  their  ages,  the  elder  says  to 
the  younger,  "  3  years  ago,  my  age  was  to  yours  as  4  to  3 ; 
but  3  years  hence,  if  we  live,  my  age  will  be  to  yours  as 
6  to  5.'*     Required  their  ages  at  the  time. 


§22.]      CONTAINING    TWO    UNKNOWN    Q,UANT1TIES.       117 

7.  A  laborer  agreed  to  reap  6  acres  of  wheat  and  5 
acres  of  rye  for  $  28 ;  but  after  he  had  reaped  4  acres  of 
the  wheat,  and  3  acres  of  the  rye,  he  was  taken  sick,  and 
received  for  the  work  he  had  done  $18.  Required  the 
price  of  reaping  an  acre  of  each. 

8.  A  jockey  says,  "  ^  of  the  worth  of  my  horse,  and  § 
of  the  worth  of  my  saddle,  make  $60;  also,  -fiy  of  the 
worth  of  my  horse,  and  f  of  the  worth  of  my  saddle, 
make  $42.''     Required  the  estimated  value  of  each 

9.  A  fishing-rod  consists  of  two  parts,  such  that  the 
lower  part,  added  to  ^  of  the  upper  part,  makes  20  feet ; 
moreover,  5  times  the  lower  part,  added  to  3  times  the 
upper  part,  exceeds  twice  the  whole  length  of  the  rod  by 
65  feet.     Required  the  length  of  each  part. 

10.  In  a  certain  school,  J  of  the  number  of  boys  ex- 
ceeds f  of  the  number  of  girls  by  19 ;  and  ^  of  the  num- 
ber of  girls,  together  with  f  of  the  number  of  boys,  makes 
62.     Required  the  number  of  each. 

11.  After  A  had  gained  $100,  and  B  had  lost  $50, 
they  had  equal  sums  of  money;  but  if  A  had  lost  $50, 
and  B  had  gained  $50,  B  would  have  had  twice  as  much 
money  as  A.     How  much  money  had  each '? 

Art.  74:.  The  following  problems  are  intended  to 
exercise  the  learner  in  the  three  different  modes  of  elimi- 
nation. Sometimes  one  mode,  sometimes  another,  will  be 
most  convenient.  It  is  advisable,  however,  that  the  pupil 
perform  each  question  of  this  article  in  the  three  different 
wa^s,  in  order  to  acquire  skill  in  eliminating. 

The  sum  of  two  numbers  is  to  their  difference  as  4 
is  to  1 ;  moreover,  the  sum  of  twice  the  greater,  and  3 
times  the  less,  is  190.     Required  the  numbers. 

2.  The  sum  of  two  numbers  is  12,  and  their  difference 
is  2.     Required  these  numbers. 


118  EQUATIONS    OF    THE    FIRST    DEGREE  [<§>  22. 

3.  A  grocer,  having  two  casks  of  wine,  drew  9  gallons 
from  the  greater  and  6  gallons  from  the  less,  and  found 
the  number  of  gallons  remaining  in  the  greater  to  the 
number  remaining  in  the  less  as  9  is  to  5.  He  then  puts 
6  gallons  of  water  into  the  greater,  and  5  into  the  less, 
and  finds  the  number  of  gallons  of  liquor  in  the  greater 
to  the  number  in  the  less  as  12  to  7.  How  many  gallons 
of  wine  were  there  at  first  in  each  ? 

1.  A's  money  and  ^  of  B's  make  $30 ;  and  B's  money, 
with  -^  of  A's,  makes  $35.     How  much  has  each? 

5.  Three  men  and  4  boys  earn  in  a  day  £1  6  s.;  also, 
5  men  and  7  boys  earn  in  a  day  c£2  4  s.  Required  the 
daily  wages  of  a  man  and  a  boy  respectively. 

6.  There  are  two  numbers,  such  that,  if  f  of  the  greater 
be  subtracted  from  the  less,  the  remainder  will  be  16;  but 
if  I  of  the  less  be  added  to  the  greater,  the  sum  will  be 
91.     Required  the  numbers. 

7.  I  hired  a  horse  for  a  journey  of  20  miles,  and  a 
chaise  for  a  journey  of  15  miles,  for  $  2*65  ;  but,  chang- 
ing my  first  intention,  I  rode  in  the  chaise  only  10  miles, 
and  went  a  distance  of  25  miles  with  the  horse,  and  paid 
for  both  $2*70.  Required  the  price  of  the  horse  and 
chaise,  respectively,  per  mile. 

8.  A  farmer  found  that  3  horses  and  4  cows  would, 
during  the  winter,  consume  12  tons  of  hay ;  and  that  it 
required  20J  tons  to  keep  5  horses  and  7  cows  the  same 
time.  Required  the  quantity  of  hay  eaten  by  a  cow  and 
a  horse,  respectively,  during  the  winter. 

9.  The  sum  of  the  distances  passed  over  by  two  loco- 
motives, the  first  running  6  hours,  and  the  second  7  hours, 
is  290  miles ;  but  the  first  goes  35  miles  more  in  3  hours 
than  the  second  does  in  2  hours.  Required  the  distance 
each  goes  per  hour. 

10.  After  A  had  lent  B  $5,  he  had  ^  as  much  money 


"-% 


/9^ 


<§>  23.]     CONTAINING  THREE  UNKNOWN  QUANTITIES.      119 

as  B ;  but  if  B  had  lent  A  f  5,  each  would  have  had  the 
same  sum.     How  much  money  had  each  ? 

11.  Ten  boxes  of  raisins  and  3  barrels  of  flour  weigh 
850  lbs. ;  also,  12  boxes  of  raisins  and  7  barrels  of  flour 
weigh  1700  lbs.  Required  the  weight  of  a  box  of  raisins, 
also  that  of  a  barrel  of  flour. 

^  12.  What  fraction  is  that,  to  the  numerator  of  which 
if  1  be  added,  the  value  of  the  fraction  will  be  f ,  but  if 
3  be  subtracted  from  the  denominator,  the  value  of  the 
fraction  will  be  f  ?  • 


SECTION   XXIII. 


EQUATIONS     OF     THE     FIRST     DEGREE     CONTAINING     THREE 
UNKNOWN     QUANTITIES. 


Art.  75 •  1.  A  boy  bought  an  apple,  an  orange,  and  a 
peach,  for  6  cents ;  2  apples,  3  oranges,  and  4  peaches,  for 
19  cents ;  5  apples,  2  oranges,  and  7  peaches,  for  25  cents. 
What  did  he  pay  for  one  of  each  ? 

Let  X  ==  the  price  of  an  apple,  y  =:  that  of  an  orange,  and 
z  =■  that  of  a  peach,  in  cents.     Then, 

(1)  x+    y+    z=    6;) 

(2)  2x  +  33/  +  4z=:19;  V 

(3)  5x  +  2y  +  7z=:25.  ) 

In  this  problem,  we  have  three  independent  equations, 
containing  three  unknown  quantities.  Our  first  step,  in 
the  solution,  is  to  obtain  from  these  original  equations  two 
others,  which  shall  contain  only  two  unknown  quantities. 
Let  us  eliminate  x,  that  is,  find  two  equations  which  shall 
not  contain  x. 


120  EQ;UATIONS    OF    THE    FIRST    DEGREE  [<§)  23. 

Mr  St  method.  Bring  down  the  2d,  and  multiply  the 
1st  by  2,  so  as  to  make  the  coefficient  of  x  the  same  as  it 
is  in  2d , 

(2)  22;  +  3y  +  4z+19;  '' 

(4)  2a;4-2y-f-22;=12.    Subtract  4th  from  2d, 

(5)  v-^^z  —  1    [  Multiply  1st  by  5,  and  bring 
^  ^   ^'^  \  down  3d, 

(6)  5a;  +  5y  +  5;^=:30; 

(3)  5a:  +  2y  +  '^2;z=25.    Subtract  3d  from  6th, 
...  (7)   3y—  22;  =  5.  )  Bring  down  5th, 

(5)  y+^z=l.S 
The  5th  and  7th  contain  only  the  two  unknown  quanti- 
ties y  and  z.  These  equations  may,  therefore,  be  solved 
as  we  solved  similar  equations  in  the  preceding  section. 
Since  the  coefficients  of  z  are  alike  in  the  5th  and  7th, 
by  adding  these  equations,  we  have 

4  y  =  12  ;  .•.  y  =  3  cents,  price  of  an  orange. 

Substitute  3  for  y  in  the  5th,  which  contains  only  y  and  z, 

3  -(-  2  2;  =  7,  .•.  2;  z=  2  cents,  price  of  a  peach. 

Substitute  3  for  y,  and  2  for  z,  in  the  1st, 

X  -f-  3  +  2  =  6,  .*.  X  =  1  cent,  price  of  an  apple. 

Second  method.     Resume  the  original  equations, 

(1)  ^+    y+    ^=    6;) 

(2)  2x  +  33^  +  4zz=19;  > 

(3)  5x-|-2y  +  ^^  =  25.  3 

Find  the  value  of  x  from  each  of  the  equations,  as  if  y 
and  z  were  known. 

(4)  xz:=6  —  y  —  z\ 

(5)  .  =  i^|=i-^ 

(6)  ^^?5:zlplf. 

Put  equal  to  each  other  the  values  of  x  in  the  4th  and 
5th ;  also  the  values  of  x  in  the  4th  and  6th, 


^23.]    CONTAINING  THREE  UNKNOWN  QUANTITIES.      121 

(7)  6  —  y  —  z  = -| ; 

(8)  6  —  y  —  z  = ^ . 

Clear  the  7th  and  8th  of  fractions ;  then  transpose  the 
terms  containing  y  into  the  first  members,  all  the  other 
terms  into  the  second,  and  reduce, 

(9)  y  =       7  —  Sz;  )  Change  the  signs  in  10th,  and 
(10)   — 3y=:  — 5  — 2z.  )  find  the  value  of  y, 
y^|v               __5-f  2z  (  Equalize  the  values  of  y  in  9th 
^     '           ^"^      3     •  I  and  11th, 

^-i^  =  7  — 22;,.-.5  +  22;=:21— 6«,.-. 

o 

z  zz:  2  cents,  price  of  a  peach. 

Substitute  2  for  z  in  9th, 

y  =:  7  —  4  =  3  cents,  price  of  an  orange. 

Substitute  3  for  y,  and  2  for  z,  in  4th, 

x=zd  —  3  —  2=1  cent,  price  of  an  apple. 

Third  method.     Take  the  original  equations, 

(1)  x+    y+    z=    6n 

(2)  2x  +  3y  +  4zz=zl9,! 

(3)  5x  +  2y  +  72;  =  25.   )Find  the  value  of  a;  from 
the  1st,  as  if  y  and  z  were  known. 

(4)  x=i6  —  y  —  z.    Substitute  this  value  in  2d  and  3d, 

(5)  12-2y-2;.  +  3y  +  4^==19;) 

(6)  30  — 5y  — 5z  +  2y-f-7%  =  25.  )  Transpose    the 
known  terms,  and  reduce  in  the  5th  and  6th, 

(7)  y  +  2z=       7;) 

(8)  —  3y  +  2  2:=:  — 5.  )  Find  the  value  of  z  in  7th, 

(9)  z  =1  -^.     Substitute  this  value  in  8th, 

-3y  +  7  — y  =  — 5,.-.  — 4yz=z— 12,.-.4y=12,  and 

y  =:  3  cents,  price  of  an  orange. 

Substitute  3  for  y  in  9th, 


122  EQ,UATIONS    OF    THE    FIRST    DEGREE  [^23 

z  =  -^^  =  2  cents,  price  of  a  peach. 

Substitute  3  for  y,  and  2  for  z,  in  4th, 
a:zz:6  —  3  —  2=  1  cent,  price  of  an  apple. 
2.    A  farmer  found  that  the  number  of  his  sheep  ex^ 
ceeded  by  20  that  of  his  cows  and  horses  together ;  that 
his  horses  and  ^  of  his  cows  equalled  in  number  -J  of  his 
sheep ;  and  that  ^  of  his  cows,  ^  of  his  horses,  and  -J  of 
his  sheep,  made  12.     Required  the  number  of  each. 
Suppose  he  had  x  sheep,  y  cows,  and  z  horses.     Then 
(1)    x  =  y  +  z-\-m; 

(2)  ^+i  =  |; 
(^)  l+T+f  =  12- 

First  method    Remove  the  denominators,  and  transpose 
the  unknown  terms  into  the  first  members. 

(4)  x  —  y  —  z=   20;  ^ 

(5)  _4a;4-5y-(_20zz:=      0;> 

(6)  8x  +  53/  +  20^  =  480.   ) 

Let  us  eliminate  y.     Multiply  4th  by  5, 

(7)  ^x  —  5y  —  5z=  100.     Add  5th  and  7th,  also  6th 
and  7th, 

(8)  2;  +  15zz=100;) 
'(9)    13  X  4- 15  2;  =  580.  / 

Since,  in  the  8th  and  9th,  the  coefficients  of  z  are  alike, 
by  subtracting  the  8th  from  the  9th,  we  have 
12  x  =  480, .-.  2:  ziz  40  sheep. 
Substitute  40  for  x  in  8th, 
40  +  15  2;  zz:  100, .-.  z  =  4  horses. 
Substitute  40  for  x,  and  4  for  z,  in  4th, 
40  —  y  —  4  =  20, .-.  y  =  16  cows. 
Second  method.     Resume    the   original    equations,   in- 
verting the  2d, 


§23.]    CONTAINING  THREE  UNKNOWN  QUANTITIES.      123 


(1)  X  =y+z+20; 

(2)  T=-+f: 


5  '     4  ■  > 

'-  =  12.) 


(3)  ^  +  -^  + 

Let  us  eliminate  x.     Bring  down  the  1st, 
(1)    a:z=:y-(-2-f-20.     From  2d  we  have 

(4)  ^^?^£±^^;from3d 

(5)  x  =  ^g . 

Equalizing  the  values  of  x  in  1st  and  4th,  also  the 
^ralues  in  1st  and  5th, 

(6)  ^I±Il  =  ,  +  .  +  20; 

(7)  y  +  ^  +  20  =  ^:^=^. 

The  6th  and  7th  do  not  contain  x.  From  these  elimi- 
nate y.  Multiply,  transpose,  and  deduce  the  value  of  y 
from  each.     The  6th  gives 

(8)  yizrSO  — 16z.  ^  FromTth, 

/p.         320  — 28g       ? 

^  f   y  13      *    /  Equalizing  the  values  of  y  in  8th 

and  9th, 

— ^ =  80  — 16  2;,  from  which  we  have 

z  izr  4  horses. 

Substitute  4  for  z  in  8th 

y  zz:  80  —  64  =:  16  cows. 

Substitute  16  for  y,  and  4  for  2,  in  1st, 

:Kn:16  +  4  +  20zz:40  sheep. 

Third  method.     Bring  down  1st,  and  clear  2d  and  3d 

from  fractions, 

(1)    x  =  y  +  z  +  20;  ^ 

(4)    4:c=::202;  +  5y;  > 

^5)    8a;  +  5y  +  20z=:480.  ) 


124  EQUATIONS    OF    THE    FIRST    DEGREE  [^23 

Let  us  eliminate  x.     Substitute,  in  4th  and  5th,  the 
value  of  X  given  in  1st, 

(6)  4y  +  4z  +  80z=202  +  5y;  ) 

(7)  8  y  +  8  z  +  160  +  5  y  +  20  z  =  480.  /  Transpose 
and  reduce  in  6th  and  7th, 

(8)  y  +  lQz=    80;  ) 

(9)  13y  +  28z  =  320.  I 

Now  eliminate  y  from  8th  and  9th.     The  8th  gives 
(10)    y  HZ  80  —  16  z.     Substitute  this  value  of  y  in  9th, 
1040  —  208  z  +  28  z  =  320, .-.  —  180  z  =z  —  720, 
.-.  180z  =  720, .-. 
2  =  4  horses. 
Substitute  4  for  z  in  10th, 
y~80  — 64=  16  cows. 
Substitute  16  for  y,  and  4  for  z,  in  1st, 
x  z=  16  -  j-  4  +  20  :=  40  sheep. 
3.    The  ages  of  3  children  are  as  follows.     Four  times 
A*s  age  and  3  times  B's  make  27  years ;    3  times  A's 
added  to  C's  make  15  years ;  and  4  times  B's  with  twice 
C's  make  32  yeare.     Required  the  age  of  each. 
Let  X,  y,  and  z  (years),  represent  the  respective  ages  of 
A,  B,  and  C.     Then, 

(1)  4x  +  3y  =  27;^ 

(2)  3x+    z  =  15;  > 

(3)  4y  +  2  2;  =  32.  ) 

In  this  example,  each  equation  does  not  contain  all  the 
unknown  quantities ;  nor  is  that  necessary  in  order  that 
the  solution  may  be  possible.  Let  us  eliminate  z  by  the 
first  method.  But  since  the  1st  does  not  contain  z,  we 
have  to  eliminate  that  letter  from  the  2d  and  3d  only 
Divide  3d  by  2, 

(4)  2'y-f    z  =  \Q,    Subtract  2d  from  4th, 

(5)  2y  —  3x1=    1.  )  Bring  down  1st, 
(1)   4x  +  3y=:27.  ) 


^23.]    CONTAINING  THREE  UNKNOWN  QUANTITIES.      125 

Let  us  now  eliminate  y  from  5th  and  1st.  Multiply  5th 
by  3,  and  1st  by  2, 

(6)  6y  — 9xz=    3;  ) 

(7)  8  X  +  6  y  =  54.  )  Subtract  6th  from  7th, 

17  x  z=  51,  .*.  X  :=  3  years,  A's  age. 

Substitute  3  for  x  in  1st, 

12  +  3  y  z=  27, .-.  y  =  5  years,  B's  age. 

Substitute  3  for  x  in  2d, 

9  -f-  z  =1 15, .-.  z  =1  6  years,  B's  age. 

Let  the  learner  solve  this  question  according  to  the 
second  and  third  methods 

From  what  precedes,  it  is  evident  that  the  three  modes 
of  elimination,  given  in  the  last  section,  may  be  ex- 
tended to  any  number  of  equations,  provided  the  number 
of  unknown  quantities  does  not  exceed  the  number  of 
equations. 

The  first  method  is  applied  to  several  equations  by 
operating  upon  these  equations  taken  two  and  two. 

In  applying  the  second  method  to  several  equations, 
find,  from  each  equation  that  contains  it,  the  value  of 
the  unknown  quantity  to  be  eliminated,  then  put  any" 
two  of  these  expressions  for  its  value  equal  to  each 
other. 

To  extend  the  third  method,  we  must,  after  having 
found,  from  one  of  the  equations,  the  value  of  the  un- 
known quantity  to  be  eliminated,  substitute  this  value 
m  every  other  equation  that  contains  this  unknown 
quantity. 

If  a  question  involves  four  unknown  quantities,  and 
gives  rise  to  four  independent  equations,  we  first  deduce 
from  them  three  equations  with  only  three  unknown  quan- 
tities ;  we  then  proceed  with  these  three  equations,  as  we 
have  done  with  the  preceding  equations  in  this  section. 

If  either  of  the  equations  does  not  contain  the  unknown 


126  SUBSTITUTION    OF    NUMBERS  ["§>  24. 

quantity  to  be  eliminated,  that  equation  may  be  put  aside 
to  be  placed  in  the  next  set  of  equations,  viz.,  those  which 
contain  one  less  unknown  quantity. 

Either  method  of  elimination  may  be  used,  but  the  first 
will  generally  be  found  the  most  convenient,  because  it 
does  not  give  rise  to  fractions.  The  pupil  is  advised, 
however,  t3  perform  each  problem  in  the  three  ways,  in 
order  to  acquire  skill  and  be  able  to  judge  which  will  be 
best  in  any  particular  case.  It  is  not  necessary  that  the 
same  mode  of  elimination  be  pursued  throughout  the  solu- 
tion of  a  question,  but  either  may  be  resorted  to  whenever 
it  shall  seem  the  most  convenient. 

Let  the  learner  find  the  values  of  x,  y,  and  z,  in  the 
following  sets  of  equations. 

4.  a;  +  yz=l3;^  6.  x-f  y=z:90—  z;^ 
x-{-z=zU'S  2x  +  40z=3y  +  20;( 
y-\-z=l5.)                   \  2;4-20i=22;-f   5-  )         ' 

5.  2x-\-Si/  +  ^zz=29n     7.   ±x-\-i^  +  iz  =  62;^         < 
3a;  +  2y  +  5z=i32;>    \       ^a;  +  |y +  ^z  =  47  ;  V 

4:^  +  3^  +  2^  =  25.)    I       ^x^ '^y  +  iz=zS8.)     ^ 


SECTION   XXIV. 

SUBSTITUTION    OF    NUMBERS    IN    ALGEBRAIC    QUANTITIES. 

Art.  70.     Let  the  learner  find  the  numerical  value  ol 
the  following  expressions,  when  a  =  4,  6  r=  3,  c=:2,  and 

1.  abc,       Ans.  4.3.2  =  24. 

2.  a^b.         Ans.  42.3  =  16.  3  =  48. 

3.  62. c3.      Ans.  32.23  =  9.8  =  72. 


^24.]  IN    ALGEBRAIC    Q,UANTTTIES.  121 

A:,    ahd.  \^.  ahc-\-d, 

5.  a^d^.  20.  a^Jy^  +  c^d. 

6.  ahc^d.  21.  ah  +  c  —  d, 

7.  ahc^d.  22.  a6c2  — ^2. 

8.  62^3.  23.  a^  —  c-\-d, 

o     «       A       4.--0  ^^-  2a  +  3c  — 26. 

9.  -.     Ans.f^3.  ^^  5«_262  +  c3. 

a26  26.  3a2  +  462_c_2rf. 

cd*  27.  («  +  6)c. 

^  28.  ac{h  +  dy 

a  '  29.  (a  +  6)(c  +  t/). 

a^c  30.  (a-f^)(^— 4 

d  •  31.  {2a  —  b){Sc  +  ^d). 

g     ^«  32.  {b^-{-a^){d—c). 

'     63-  33.  (3a2_262)(4^_3c). 

^  34.  (a2  — 62)(c2_e^2). 

•    ac*  35.  a^{Sb  —  c  +  2d), 

J.     ^c  36.  (62  — a2)(j2_c2). 

'      a3-  37.  3c2(a-|-^_J). 

16.  a  +  6  +  c.     Ans.  9.        38.  2  6^  (a  +  c^  +  </). 

17.  a^-^b-{-d,  39.  6«(a2_6-f  c). 

18.  ab  +  cd.  40.  «(262_c2  +  ^3). 

Find   the   value   of   the   following   expressions,   when 
=  5,  6  =  3,  m  =  2,  and  n=zO. 

41.  «-|-6  —  m-|-w. 

42.  3«  —  5w-(-4m. 

43.  ab-\-mn. 

44.  67wa  —  5abn. 

45.  m7i-|-3a6n.  ^j 

46.  4a6c7i  —  5a 6^. 

47    ^±^.  52. 

m  —  n  3  m2  — .  n^ 


49. 
50. 


t3mn 
26  * 

3(a- 

-6)  (m  —  n) 

m 

(36- 

.2«)m« 

2771 

—  3n 

2(a2- 

-62)(a«4-62) 

48 


3a  — 6 
m2 


128  SUBSTITUTION    OF    NUMBERS,    &C.  [<§>  24. 

Substitute  numbers  in  the  following  equivalent  expres- 
sions, showing  their  identity,  whatever  numbers  are  put 
instead  of  the  letters;  provided,  however,  that  the  same 
value  be  given  to  the  same  letter  in  the  two  members  of 
an  equation.  An  identical  equation  is  one  in  which  the 
two  members  are  exactly  alike ;  they  may  differ  in  fornix 
but  both  can  be  reduced  to  the  same  form. 

53.  (a  +  c)6  =  «6  +  Z>c. 

In  this  example,  let  a:=z:l,  6  =  2,  and  c=:3.  Then 
the  first  member  becomes  (1  +  3)2  =  4.2  =  8.  The 
second  member  gives  1.2-|-2.3  =  2-f-6  =  8.  The  re- 
sult B,  we  perceive,  are  alike. 

54.  am  —  bm=:m{a  —  b). 

55.  {a  —  m)  {a -\- m)  =1  a^  —  m^. 

56.  {a-\-m){a-{'m)  =  a^'^2a?n'{-m^. 

57.  {a  —  m){a  —  jn)=za^  —  2a7n-\-  m^, 

58.  "-±^-^."-=^  =  a. 


o— 1 


62.    "^^^=nfi-\-l. 

m^  —  l  ' 

a-\-  m 
64.    (a  +  m)  (a-|~n)  =  a2-|-a(m-[-n)  +  m». 
b        ,    a  —  b  cP- 


65 
66. 


a  +  6'        b  ab  +  ¥ 

a-\-m    ^^a^m 2  (a?  +  m^) 

a  —  m        a-\-7n  a^  —  m2 


^25.]  LITERAL    EQUATIONS.  *  "     129 


67. 


fl?^10a+9        a  — 9* 


^      (n+l)(n+2)_n(n+1)    i    ^^    I    ^ 
2  2  '         '"    ' 


SECTION  XXV. 

LITERAL    EQUATIONS. 

Art.  77.      Find   the  values   of  x   in   the   following 
equations. 

1.    f^±^  =  a:  +  3a.     Multiply  by  6, 

a'\-mxz=zhx'\-^ah.     Transpose  a  and  6 x, 
mx  —  bxzzzSab  —  «.     Separate  the  first  member  into 
factors,  one  of  which  is  x, 
{m  —  b)xz=Sab  —  a. 
In  this  equation,  x  is  taken  m  —  b  times,  that  is,  thf* 
factor  m  —  6  is  the  coefficient  of  x.     Divide  both  mem- 
bers by  this  coefficient, 

S  ab  *—  a 


m  —  b 

2.    i^±^^^nA5  — 2a;_6.     Multiply  by  a +  «, 

Sc-\-ax  —  bxzzn^ax  —  ab-\-2mx  —  bm. 
Transpose  the  terms  containing  x  into  the  first  member, 
the  others  into  the  second,  and  reduce, 
—  ax  —  bx  —  2mxz=z  —  ab  —  bm  —  3c.     Change  the 

signs, 

ax-{-bx-{-2mx=zab'{-bm-{-Sc,     Separate  the  first 

member  into  factors,  one  of  which  is  x. 


130  *  GENERALIZATION.  [<§»  26 

{a-{-b-\-2m)xi=zab  -{-bm-^-^c.     Hence 

_  a  6  +  bm-{-3c 


SECTION  XXVI. 


GENERALIZATION. 


Art.  78.  In  pure  algebra,  letters  are  used  to  repre- 
sent known  as  well  as  unknown  quantities.  It  enables  us 
to  conduct  operations  with  greater  facility,  and  to  form 
rules.  The  results  of  purely  algebraical  solutions  of 
problems  are  called  general  for  mulcB,  The  design  of  this 
section  is  to  familiarize  the  learner  with  such  solutions. 

1.  A  and  B  have  together  $270,  but  B  has  twice  as 
much  money  as  A^,     Required  the  money  of  each. 

In  this  problem  it"  is  required  to  divide  $270  into  two 
parts,  one  of  which  shall  be  double  the  other. 

Let  X  dollars  be  A's  share ;  then  2  x  dollars  will  be  the 

share  of  B.     Hence, 

X  -j-  2  X  zzi  270.     This  equation  gives 

xziz    $90,  A's  share;  )  ^^ 

2  X  =  $  180,  B's  share ;  i 

Now,  suppose  that,  instead  of  $270,  A  and  B  have  any 

number  a  of  dollars,  of  which  B  has  twice  as  much  as  A. 


§  26.]  GENERALIZATION.  181 

In  this  case,  the  problem  is  to  divide  the  number  a  into 
two  parts,  one  of  which  shall  be  twice  as  great  as  the 
other. 

Representing  the  shares  as  before,  we  have 

This  equation  gives 

X  :=  — ,  A's  share  ;  ) 

2  /  Ans,     General  formulae. 
2  2;=:  — ,  B's  share  ;  \ 

The  general  formulae  show  us  that  one  part  is  -J-  and 
the  other  f  of  the  number  to  be  divided,  without  reference 
to  the  particular  value  of  that  number. 

If,  in  the  general  formulae,  we  now  put  $  270  instead 
of  a,  we  have 

x=f  :=  190,  A's  share;  ) Particular 

2z:^'-^°  =  2.90=  f  180,  B's  shared  '"'^"■'- 

3  / 

Let  the  learner  substitute  other  numbers  for  a  in  the 
general  formulae,  and  find  the  particular  answers.  Any 
number  divisible  by  3  will  give  whole  numbers  for  the 
answers. 

2.  The  sum  of  the  ages  of  two  brothers  is  76  years, 
and  the  difference  of  their  ages  is  16  years.  Required 
the  age  of  each. 

In  this  problem  we  are  to  separate  76  into  two  parts, 
such  that  the  difference  of  those  parts  shall  be  16.     In- 
stead of  76,  let  us  suppose  that  the  number  to  be  sepa- 
rated into  parts  is  indefinite,  and  that  it  is  represented  by 
a ;  also,  that  d  represents  the  difference  of  the  parts,  that 
is,  the  excess  of  the  greater  part  above  the  less. 
Let  X  =  the  less  part ;  then 
2  +  c?  :=  the  greater  part.     Hence, 
X -^^  X -^^  d  z=z  a.     Reduce  and  transpose  c?, 


>'^       V 


^"-t^^  u^--  ^-:f+  1'' 


'^v  ^''*-'  yfl"-  '"i+  '^    ''j'ti- 

132  GENERALIZATION.  [<§>  26. 

2x=ia  —  d;  divide  by  2, 
x=i V  ^^  ~ — '  ^^^  ^^^^  i?«r^.   (Art.  67.) 

To  obtain  the  greater  part,  we  add  d  to  the  less,  and 
vie  have 

x-^-dz^-—  —  -^H-^-     Change  d  in  the  2d  member  to 

halves, 

+    -,         a          d     .    2  d  , 

a  =  — ;  reduce, 

x-^dzzz 1 := ,  the  greater  part. 

By  examining  the  general  formulaB  for  the  two  parts, 
and  recollecting  that  a  and  d  may  stand  for  any  numbers, 
provided  d  is  less  than  «,  we  see  that  they  furnish  the  fol- 
lowing rule  for  separating  a  number  into  two  parts,  whose 
difference  is  given. 

The  less  part  is  found  by  subtracting  half  of  the  dif- 
firence  of  the  parts  from  half  of  their  sum,  that  is,  from 
half  of  the  number  to  be  separated ;  or,  by  subtracting  the 
difference  of  the  parts  from  their  sum,  and  dividing  the 
remainder  by  2. 

The  greater  part  is  found  by  adding  half  of  the  dif 
ference  to  half  of  the  sum  of  the  parts ;  or,  by  adding 
the  difference  to  the  sum  of  the  parts,  and  dividing  the 
amount  by  2. 

Let  the  learner  separate  the  following  numbers  into 
parts,  either  by  means  of  the  formulae,  or  by  following 
the  rule. 

\j^^'     JSTumbei's  to                                                           Differences 
^    4^   be  separated.                                                        of  the  Parts. 
d  50 10. 

4.  30  .     .     .  16. 

5.  100 20. 

6.  35 5 


§)  26.J  GENERALIZATION.  133 

JVumbers  to  Differences 

be  separated.  of  the  Parts. 

7,         106      ..     ,  50. 

a  50 5. 

9.  33      . 8. 

10.  A  man  bought  corn  at  a  shillings,  rye  at  b  shil- 
lings, and  wheat  at  c  shillings,  per  bushel,  and  the  whole 
amounted  to  d  shillings.  Required  the  number  of  bushels 
of  each  he  bought,  if  he  bought  the  same  number  of 
bushels  of  each. 

Let  X  =z  the  number  of  bushels  of  each.     Then, 

ax-\-bx-^cx=:d. 

Separating  the   1st  member  into  factors,  one  of  which 

IS  X,   (Art.  59.) 
{a -f- b -{- c)  X  =^  d ;  dividing  by  a  -|-  6  +  c,  the  coefficient 

of  X, 

X  =: ,  the  number  of  bushels  of  each. 

a+6  +  c 

This  general  formula  may  be  translated  into  the  follow- 
ing rule.  \ 

Divide  the  price  of  the  whole  by  the  sum  of  the  prices 
of  a  bushel  of  each  sort ;  the  quotient  will  be  the  number 
of  busJiels  of  each. 

If,  in  the  formula,  we  substitute  5,  6,  7,  and  180,  for  a, 
6,  c,  and  c?,  respectively,  we  have 

X  r^ =  —  zz:  10  bushels,  particular  answer. 

5  +  6-1-7         18  '^ 

This  rule  may  be  extended  to  any  number  of  articles ; 
only  we  must  change  *'  bushel  "  and  "  number  of  bushels  " 
into  other  corresponding  expressions,  as  the  case  may 
require, 

11.  How  many  apples  at  1  cent,  pears  at  2  cents, 
peaches  at  3  cents,  and  oranges  at  4  cents,  apiece,  of 
each  an  equal  number,  can  I  buy  for  $5? 

12.  In   a  certain  manufactory  there   are  employed  6 


134  GENERALIZATION.  [^  26  : 

times  as  many  boys  as  men,  and  c  times  as  many  girls  as 
boys.  Required  the  number  of  each,  the  whole  numbei 
of  individuals  being  a. 

Let  X  z=  the  number  of  men ; 

then  bx=z  the  number  of  boys, 

and  bcx=^ the  number  of  girls.     Hence, 

x-^bx-^bcxz=za. 

Here  x  is  taken  l-\-b-\-b  c  times ;  therefore,  separating 

the  first  member  into  factors,  one  of  which  is  x, 

(l  +  6  +  6c)a:  =  a.     Divide  by  1  +  6  +  6  c, 


a 


l  +  b  +  bc 


,  number  of  men : 


bx=i ; X  b,  number  of  boys : 

X  b  c,  number  of  girls. 


l-\-b  +  bc 

If,  m  these  formulae,  we  substitute  81  for  a,  2  for  6, 
and  3  for  c,  we  have 

Of  01 

— =  — =:9men;  9.2=  18  boys,  and  9.2.  3  =  54 

1  +  2+6         9  -^ 

girls. 

13.  What  will  be  the  particular  answers  in  the  last 
question,  if  a  is  144,  b  is  5,  and  c  is  6? 

14.  Two  men.  A'  and  B,  engaged  to  dig  a  well;  A 
could  dig  it  alone  in  a  days,  and  B  could  dig  it  alone  in  b 
days.     In  what  time  could  they  both  together  dig  it  \ 

Let  X  =  the  number  of  days  required.     Then,  as  A 

could  dig  the  whole  in  a  days,  in  1  day  he  would  dig  —  of 

a 

it ;  and,  as  B  could  dig  the  whole  in  b  days,  in  1  day  he 
would  dig  —  of  it.    Both  would,  therefore,  dig 1 of 

it  in  1  day ;  and  in  x  days  they  would  dig 1 of  it. 

But  we  have  supposed  that  in  x  days  they  would  dig  the 
whole.     Hence, 


'^  26.]  GENERALIZATION.  136 

T +  1=1. -ell. 

Multiply  by  a  and  6, 

h%-\-  axz=.  ah\  or, 

(6  +  «)  x  zz:  a  6 ;  divide  by  6  -(-  «, 

ah        . 

X  -=. ,  Ans. 

6+a' 

If  a  is  6,  and  h  is  7,  we  have 
2:  =  --ij~7 1~  ^f  ==  3^3^  days,  particular  answer. 

From  the  general  formula  in  this  question,  we  derive 
the  following  rule  for  any  case  in  which  there  are  but 
two  workmen. 

Divide  the  product  of  the  numbers  expressing  the  times 
in  which  each  would  perform  the  work,  by  the  sum  of  those 
numbers.  The  quotient  will  be  the  time  in  which  both 
together  will  perform  it. 

Let  the  two  following  questions  be  performed  by  the 
preceding  rule. 

15.  A  man  could  do  a  piece  ol  work  in  6  days,  and  a 
boy  could  do  the  same  in  10  days.  In  what  time  would 
they  both  together  do  it? 

16.  A  man  alone  would  consume  a  barrel  of  flour  in  6 
months,  and  the  same  quantity  would  last  his  wife  8 
months.     How  long  would  a  barrel  supply  both? 

17.  Let  it  be  required  to  find  a  rule  for  dividing  the 
gain  or  loss  in  partnership,  or,  as  it  is  commonly  called, 

he  rule  of  fellowship.      Let    us  first  take   a  particular 
example. 

Three  men,  A,  B,  and  C,  traded  together,  and  gained 
$300.  A  put  into  the  partnership  $7  as  often  as  B  put 
in  $6,  and  as  often  as  C  put  in  $2.  Required  each 
man's  share  of  the  gain. 


136  GENERALIZATIOM.  [<^  26.|f^ 

Let  X  =1  A's  share.  B  put  in  f ,  and  C  put  in  f  as 
much  stock  as  A.     Therefore,  B  must  have  f ,  and  C  f-  as 

much  gain  as  A.     Consequently,  —  =  B's  share,  and  — 

=  C'fe  share.     Hence, 

G  X        2  a? 
X  +  —  -f-  —  =  300.     This  equation  gives 

X  ==  $140,  A's  share; 

—  =  — —  =  6  .  20  :=  $120,  B's  share , 

1.1 

—  =z  — —  i=:2  .  20  =  $40,  C's  share. 

7  7  ^       , 

To  generalize  this  question,  suppose  that  A  put  in  m 
dollars  as  often  as  B  put  in  n,  and  as  often  as  C  put  in  p 
dollars ;  and  that  they  gained  a  dollars.     Then  B  put  in 

n  P 

— ,  and  C  —  as  much  as  A ;  they  must,  therefore,  have, 

respectively,  —  and  —  as  much  gain  as  A. 
Let  a;  =  A's  gain  ;   then 
—  =  B's  gain,  and 

—  =  C's  gain.     Hence, 

^  +  ^  +  ^  =  ^-     Multiply  by  m, 
mX'\-nx-\-px=zma;   or, 
{m-\-n  -\-p)  x=zm  a;  divide  by  m  -f-  n  -f-p. 


ma 


,  or  x=^m  X ,  A's  share. 


m  +  w-f-p  m-^n-^-p 

B's  share  is  —  of  A's :    —  of  tw  X — 


IS 


m  +  n  +  p        m  +  w  +  l?^ 

and  —  of  it  is  n  times  as  much :   hence,  —  z=:n  X 
,  B's  share. 


^26.]  GENERALIZATION.  137 

P 

C  s  share  being  —   as  much  as  A's,  we  have,  in  like 

manner, 

^  z=  n  X ,  C's  share  ; 

If  we  recollect  that  m,  n,  and  p  are  the  proportions  of 
stock  furnished  respectively  by  A,  B,  and  C,  we  shall 
perceive  that  the  whole  gain  a  is  divided  by  m  -f-  w  +jp, 
the  sum  of  the  proportions  of  stock  furnished  by  all  the 
partners,  and  that  this  quotient  is  multiplied  by  m,  A's 
oroDortion,  by  n,  B^s  proportion,  and  by  p,  C's  proportion 
of  the  stock,  to  obtain  their  respective  shares  of  the  gain. 
Hence,  since  a  may  represent  the  loss  as  well  as  the  gain, 
for  finding  each  partner^s  share  of  gain  or  loss,  we  have 
the  following  rule. 

Divide  the  whole  gain  or  loss  hy  the  sum  of  the  propor- 
tions  of  the  stock,  and  multiply  the  quotient  by  each  part' 
ner's  proportion. 

This  rule  applies,  whatever  be  the  number  of  partners. 

18.  James  and  White  trade  together,  the  former  putting 
into  the  partnership  $800,  and  the  latter  $600.  They 
gain  $700.  Find  by  the  formulae,  or  by  the  rule,  each 
partner's  share  of  the  gain. 

Remark.  If  the  learner  use  the  formulae,  in  this  case, 
since  there  are  but  two  partners,  p  must  be  zero.  More- 
over, when  the  amounts  actually  put  in  are  given,  the 
simplest  proportions  of  the  stocks  will  be  found  by  divid- 
ing these  amounts  by  the  greatest  number  that  will  divide 
them  all  without  a  remainder.  Thus,  800  and  600  are 
both  divisible  by  200,  and  the  quotients,  4  and  3,  are  the 
simplest  proportions  of  the  stock. 

19.  What  would  be  each  man's  gain,  if  A  furnished 
$400,  B  $300,  and  C  $200,  the  whole  gain  $450? 


138  GENERALIZATION.  [§>  26. 

^y  20.  What  would  each  man  lose,  if  four  partners  fur- 
^nished,  respectively,  $700,  $600,  $400,  and  $200,  the 
whole  loss  being  $380? 

21.  A  put  in  $500  for  4  months,  B  put  in  $400  for  6 
months,  and  C  put  in  $  300  for  7  months.  They  gained 
$325.     Required  each  man's  gain. 

RemarTc.  When  the  stocks  are  employed  unequal  times, 
it  is  manifest,  that  each  partner's  stock,  or  his  proportion 
of  the  stock,  must  be  multiplied  by  the  number  expressing 
the  time  which  it  is  in  trade,  and  that  then  the  proportions 
of  these  products  must  be  used. 

Art.  79,  Generally,  known  quantities  are  represented 
by  the  first,  and  unknown  quantities  by  the  last  letters  of 
the  alphabet.  But  it  is  frequently  convenient  to  use  the 
initial  letters  of  the  names  of  quantities,  whether  known 
or  unknown. 

In  the  following  questions  of  simple  interest,  let  p  be 
the  principal,  r  the  rate,  t  the  time,  i  the  interest,  and  a 
the  amount.  It  must  be  remembered  that  r  is  supposed 
to  be  a  fraction,  as  '06,  '05,  fcc,  according  as  the  rate  is 
6  per  cent.,  5  per  cent.,  &c.,  and  that  the  time  is  ex- 
pressed in  years  and  fractions  ot  a  year. 

1.  What  is  the  simple  interest  of  pflollars,  for  t  years, 
at  r  per  cent.  ? 

Since  the  principal  multiplied  by  the  rate  gives  the  in- 
terest for  a  year,  we  have 

rpziz  the  interest  for  1  year  ;  and 

trp^z  the  interest  for  t  years.     Hence, 

i  ^=.t  r  p. 

From  this  formula  we  deduce  the  following  rule. 

To  find  the  interest,  the  principal,  time,  and  rate  being 
known,  multiple/  the  principal,  time,  and  rate  together. 


^26.]  GENERALIZATION.  139 

2.  Required  the  interest  on  $247*50,  for  5  years,  at  (> 
per  cent. 

3.  The  equation  trp=:i  contains  four  quantities,  any 
three  of  which  being  known,  the  fourth  may  be  found. 
For  example,  to  find  p  when  the  other  quantities  are 
given.     Take 

trp=zi;  divide  both  members  by  t  r, 
I 
^        tr 
This  formula  gives  the  following  rule. 

To  Jind  the  principal,  when  the  interest,  time,  and  rate 
are  known,  divide  the  interest  by  the  product  of  the  time 
and  rate. 

4.  The  interest  being  $90*10,  the  time  4  years,  and 
the  rate  5  per  cent.,  required  the  principal,  according 
to  the  rule. 

Let  the  learner  find,  from  trpz=i i,  formulae  for  the  time 
and  the  rate,  translate  these  formulae  into  rules,  and  per- 
form by  the  rules  the  particular  examples  subjoined. 

5.  The  interest,  rate,  and  principal  given,  find  the 
time. 

6.  The  interest  being  $54,  rate  6  per  cent.,  principal 
$150,  what  is  theJ;ime? 

7.  The  interest,  rate,  and  principal  given,  find  the. i  ate. 

8.  The  interest  being  $  33*705,  time  5  years,  and  prin- 
cipal $96*30,  what  is  the  rate? 

9.  Required  the  amount  of  p  dollars,  for  t  years,  at 
the  rate  r,  simple  interest. 

Since  the  amount  is  the  sum  of  principal  and  interest, 
t  rp  being  the  interest,  as  before,  we  have 

a^p-\-trp;  or,  az=:^  (1 -j- ^r). 
Hence  the  following  rule. 

To  find  the  amount,  when  the  principal,  time,  and  rate 


1 40  GENERALIZATION.  [^  26. 

are  known,  multiply  the  time  and  rate  together,  add  1  to 
the  product,  and  multiply  this  sum  by  the  principal, 

10.  The  principal  being  $630-20,  the  rate  4^  per 
cent.,  and  the  time  7  years  and  6  months,  what  is  the 
amount  ? 

11.  Let  us  find  the  value  of  p  from  the  equation, 
p-^trp^=z  a.     Separate  the  first  member  into  factors, 

{\-\'tr)p=za\  divide  by  1 -(- ^  r. 


I  \^tr 

Hence, 

To  find  the  principal,  when  the  amount,  time,  and  rate 
are  given,  multiply  the  time  and  rate  together,  add  1  to 
the  product,  and  divide  the  amount  hy  the  sum, 

12.  The  amount  being  $124,  the  rate  6  per  cent.,  and 
the  time  4  years,  what  is  the  principal  1 

Remark,  The  principal,  in  this  case,  is  called  the 
present  worth  of  the  amount. 

13.  Required  the  present  worth  of  $400,  due  in  3 
years  and  3  months,  at  5  per  cent. 

14.  From  the  same  equation,  let  us  find  the  formula 
for  r. 

p-\-trp:=ia',  transpose  p, 
trp^ziia — p',  divide  by  if |>, 

r  =1  -^^.     Hence, 
tp 

To  find  the  rate,  when  the  amount,  time,  and  principal 
are  given,  divide  the  difference  of  the  amount  and  princi^ 
pal  by  the  product  of  the  time  and  principal. 

Remark,  This  rule  is  virtually  the  same  as  that  given 
by  problem  7th,  since  the  difference  of  the  amount  and 
principal  is  the  interest. 


,3.  .^^ ' 

f  <§>  26.]  GENERALIZATION.  141 

15.  The  amount  being  $368*875,  time  3  years,  and 
principal  $325,  required  the  rate. 

16.  In  a  similar  manner,  and  from  the  same  equation, 
let  the  learner  find  the  formula  for  t,  convert  it  into  a 
rule,  and  perform  by  the  rule,  the  subjoined  particular 
example. 

17.  The  amount  being  $1012-50,  principal  $750,  and 
rate  5  per  cent.,  required  the  time. 


Art.  80.  1.  Separate  a  into  two  parts,  such  that  one 
shall  be  m  times  the  other. 

2.  A  and  B  have  together  a  dollars,  of  which  B  has  m 
times  as  much  as  A.     How  many  dollars  has  each  ? 

3.  Separate  a  into  two  parts,  such  that  the  second  shall 

be  the  —  part  of  the  first. 

4.  Separate  a  into  three  parts,  so  that  the  second  shall 
be  m  times,  and  the  third  n  times  the  first. 

5.  What  number  is  that  whose  mih  part,  added  to  its 
nth  part,  makes  the  number  a  1 

6.  What  number  is  that  whose  with  part  exceeds  its  nth 
part  by  a  ? 

7.  Separate  a  into  two  parts,  so  that  the  with  part  of 
one,  added  to  the  nth  part  of  the  other,  shall  make  the 
number  a. 

Let  X  =:  one  part ;  the  other  will  be  a  —  x, 

8.  After  paying  away  —  and  -r-  of  my  money,  I  had  c 

dollars  left.     How  many  dollars  had  I  at  first  ? 

9.  A  man  and  his  boy  together  could  do  a  piece  of 
work  in  a  days,  and  the  man  could  do  it  alone  in  b  days. 
Required  the  number  of  days  in  which  the  boy  could  do 
it  alone. 

10    The  heirs  to  an  estate  received  a  dollars  each; 


4 


142  EXTRACTION    OF    THE  [*§>  2^« 

hut  if  there  had  been  b  less  heirs,  they  would  have  re- 
ceived c  dollars  each.     Required  the  number  of  heirs. 

11.  A  man  has  4  sons,  each  of  whom  is  a  years  older 
than  his  next  younger  brother,  and  the  eldest  is  m  times 
as  old  as  the  youngest.     Required  their  ages. 

12.  A  father  gave  his  children  m  oranges  apiece,  and 
had  a  oranges  left ;  but  in  order  to  give  them  n  oranges 
apiece,  he  wanted  h  oranges  more.  How  many  children 
had  he  ? 

13.  The  sum  of  two  numbers  is   «,   and   —  of  the 

greater  added  to  —  of  the  less  makes  b.     Required  the 

numbers. 

Remark.  This  and  the  two  following  questions  may 
be  performed  by  means  of  two  unknown  quantities. 

14.  One  ox  and  m  cows  cost  a  dollars;  but  one  cow 
and  n  oxen  cost  b  dollars.  Required  the  price  of  a  cow, 
and  that  of  an  ox. 

15.  There  are  two  numbers,  such  that  the  first  with 

—  of  the  second  makes  a :  and  the  second  with  —  of  the 

m  '  n 

first  also  makes  a.     Required  the  numbers. 


SECTION   XXVII. 

EXTRACTION  OF  THE  SECOND  ROOTS  OF  NUMBERS. 

Art.  81.  1.  Some  market  women,  counting  their 
eggs,  found  that  each  had  12  times  as  many  eggs  as  there 
were  women,  and  that  they  all  together  had  300.  Re- 
quired the  number  of  women. 


<§>  27.]  SECOND    ROOTS    OF    NUMBERS.  143 

Let  X  =1  the  number  of  women  ; 

then  12  x=:  the  number  of  eggs  each  had, 

and  12  z .  x,  or  12  x^  =  the  whole  number  of  eggs.    Hence, 

12  2;2  z=  300.     Divide  by  12, 

x^,  or  x.x=z  25. 

We  see  that  x  must  be  a  number  which,  multiplied  by 

Itself,  shall  produce  25;   and  we  know  that  5.5  =  25 

Hence, 

a;  =  5  women,  Ans. 
The  equation  12  x^  =:  300  is  called  an  equation  of  the 
second  degree,  or  a  quadratic  equation,  because  it  contains 
the  second  power  or  square  of  the  unknown  quantity. 

In  general,  an  equation  of  the  second  degree  is  such 
as,  when  reduced  to  its  simplest  form,  contains  at  least 
one  term  in  which  there  are  two,  but  no  term  in  which 
there  are  more  than  two  unknown  factors. 

When  an  equation  with  one  unknown  quantity  contains 
the  second  power  only  of  that  quantity,  it  is  called  a  jpure 
equation  of  the  second  degree,  or  a  pure  quadratic  equa- 
tion.    The  equation  given  above  is  of  this  kind. 

The  first  power  of  a  quantity,  with  reference  to  the 
second,  is  called  the  second  root  or  square  root,  and 
finding  the  first  power  when  the  second  is  given,  is  called 
extracting  the  second  or  square  root.  The  second  root  of 
a  quantity,  then,  is  such  as,  being  multiplied  by  itself,  will 
produce  that  quantity. 

The  second  powers  of  the  first  nine  figures  are  as  fol- 
lows. 

i  1, 2,  3,    4,    5,    6,    7,    8,    9.     Roots. 

I  1,  4,  9,  16,  25,  36,  49,  64,  81.     Powers. 

Hence,  when  a  number  consists  of  only  one  figure,  its 

second   power    cannot   contain   more   than   two    figures. 

The  least  number  consisting  of  two  figures  or  places  is 


144  EXTRACTION    OF    THE  [*§>  27. 

10,  the  second  power  of  which  is  100,  consisting  of  three 
figures,  or  rather  of  three  places. 

It  is  to  be  observed,  that,  when  a  number  has  zeros  at 
the  right,  its  second  power  is  terminated  by  twice  as 
many  zeros ;  that  is,  the  second  power  has,  at  the  right, 
twice  as  many  zeros  as  the  root.  And,  in  general,  the 
product  of  two  or  more  numbers,  each  having  zeros  at 
the  right,  has,  at  its  right,  as  many  zeros  as  all  the  factors. 
For  example,  (30)2  =  900,  (500)2  zzi  250000,  &,c. ;  (400)2 
X  30  =:  4800000,  (2000)2 .  300  =  1200000000,  &c. 

Art.  83.  In  order  to  find  a  mode  of  extracting  the 
second  root  of  a  number  consisting  of  more  than  two 
figures,  let  us  examine  the  second  power  of  a  -f"  ^>  which 
is  a2  _|_  2  a  6  -f-  ^^>  ^^  which  a  may  represent  the  tens,  and 
b  the  units  of  the  root.  Let  a  =  2  (tens)  or  20,  and 
6  =  3;  then  «  +  6  =  23;  a2~400,  2«6=:2.20.3  = 
120,  and  62—9;  consequently,  a2_j.2a6  +  62  — 400  + 
120+9=529.     Hence, 

WTien  a  number  contains  units  and  tens,  its  second 
power  contains  the  second  power  of  the  tens,  plus  twice 
the  product  of  the  tens  by  the  units,  plus  the  second  power 
of  the  units. 

Let  us  now,  by  a  reverse  operation,  deduce  the  root 
from  the  power. 

Operation, 
5^29(23.     Root. 
4 

12'9(4.     Divisor. 
Since  the   number  contains   hundreds,  its   root  must 
necessarily  contain  tens.     Our  first  object,  then,  is  to  find 


^27.J  SECOND    ROOTS    OF    NUMBERS.  145 

the  tens  of  the  root.  But  since  the  second  power  of  the 
tens  of  the  root  cannot  have  any  significant  figure  below 
hundreds,  it  must  be  found  in  the  5  (hundreds).  We 
separate,  therefore,  the  last  two  figures  from  the  5  by  an 
accent. 

Now  the  second  power  of  2  (tens)  is  4  (hundreds),  and 
that  of  3  (tens)  is  9  (hundreds).  Therefore  the  greatest 
second  power  of  tens  found  in  5  (hundreds)  is  4  (hun- 
dreds), and  the  root  of  4  (hundreds)  is  2  (tens). 

We  place  2  at  the  right  of  the  given  number  as  the 
tens  of  the  root,  separating  it  from  that  number  by  a  line, 
and  subtract  the  second  power  4  (hundreds),  correspond- 
ing to  a^,  from  the  5  (hundreds). 

To  the  right  of  the  remainder  1  (hundred),  we  bring 
down  the  two  figures,  previously  cut  off,  and  have  129, 
which  we  call  a  dividend.  This  number  corresponds  to 
2  «  6  -|-  6^.  It  contains  twice  the  product  of  the  tens  by 
the  units,  plus  the  second  power  of  the  units.  If  it  con- 
tained 2  «  6  only,  or  exactly  twice  the  product  of  the  tens 
and  units,  we  should  find  the  units  exactly  by  dividing  by 
twice  the  tens,  for  2  a  6  divided  by  2  a  gives  6.  As  it  is, 
if  we  divide  by  twice  the  tens,  disregarding  the  remain- 
der, we  shall  obtain  the  units  of  the  root  exactly,  or  a 
number  a  little  too  great. 

Twice  2  (tens)  are  4  (tens),  by  which  we  are  to  divide. 
But  since  tens  multiplied  by  units  can  have  no  significant 
figure  below  tens,  if  we  take  4  merely  as  our  divisor,  we. 
must  reject  the  right-hand  figure  9  of  the  dividend ;  that 
is,  since  the  divisor  is  considered  as  ten  times  less  than 
its  true  value,  that  the  quotient  may  not  be.  changed,  we 
must  make  the  dividend  also  ten  times  less  than  its  true 
value.  Hence,  before  dividing,  we  separate,  by  an  accent, 
the  9  from  12. 

The  divisor  4  is  contained  3  times  in  12 ;  we  place  3 
10 


146  EXTRACTION    OF    THE  [<§>  27. 

in  the  root,  at  the  right  of  the  2,  and  obtain  for  the  entire 
root  23.  The  second  power  of  23  is  529,  showing  that 
23  is  the  exact  root  sought. 

Art  83.  In  the  preceding  example,  after  having 
fonnd  the  unit  figure,  we  raised  the  whole  root  to  the 
second  power,  in  order  to  ascertain  its  correctness.  We 
vshall  now  show  how  the  correctness  of  the  second  or  any 
succeeding  figure  of  the  root  may  be  ascertained,  without 
raising  the  whole  to  the  second  power. 

Let  us  extract  the  second  root  of  2401. 

Operation, 
24'01(49.     Root. 
16 

80'1(89 
801 

0. 

Reasoning  as  in  the  preceding  example,  we  find  the 
greatest  second  power  of  tens  contained  in  24  (hundreds), 
to  be  16  (hundreds),  the  root  of  which  is  4  (tens).  We 
put  4  as  the  first  figure  of  the  root,  subtract  its  second 
power  from  24,  bring  down,  to  the  right  of  the  remainder, 
the  two  figures  cut  off,  and  have  801  for  a  dividend. 
This  corresponds  to  ^  ah  -^h^,  which  is  the  same  as 
(2  a  -j-  5)  h.  Dividing  80  by  8,  twice  the  tens  of  the 
root,  we  have  10  for  the  quotient ;  but  since  the  unit 
figure  cannot  exceed  9,  we  put  9  in  the  root,  at  the  right 
of  4,  and  have  49  for  the  entire  root. 

Now,  in  order  to  determine  whether  9  is  the  proper  unit 
figure,  we  observe  that  our  divisor  8  corresponds  to  2  a, 
and  9  is  the  presumed  value  of  h ;  hence  80  -f-  9  or  89 
corresponds  to  2  a  -(-  5 ;  we  therefore  put  9  at  the  right 
of  the  divisor,   and   multiply  89  by  9 ;   the  product  801 


^  27.]       SECOND  ROOTS  OF  NUMBERS.  147 

answers  to  (2 « -f-  b)  6,  which,  subtracted  from  the  divi- 
dend, leaves  no  remainder.  Therefore  49  is  the  correct 
root. 

Let  the  learner  find  the  second  roots  of  the  following 

numbers,  according  to  the  method  just  explained.         ^ 

X,     1     625.'^  ^^^  5.   225.     '  ''^ 

^    2.    1089.-3-^  6.   5476.      /^^ 

J(  J.   2809.   *'^  7.   6241.       7  f 

4.    1936.    tj  ^  8.   9801.        ^  f 

Art.  84:.  Let  it  now  be  required  to  extract  the  root 
of  105625. 

Since  (10)2 z=  100,  (100)2=10000,  (1000)2^=1000000, 
it  follows  that  the  second  power  of  any  whole  number 
greater  than  10  and  less  than  100,  that  is,  consisting  of 
two  figures,  must  be  greater  than  100  and  less  than  10000, 
that  is,  it  must  consist  of  three  or  four  figures ;  also,  the 
second  power  of  any  number  between  100  and  1000,  that 
is,  consisting  of  three  figures,  must  contain  five  or  six 
figures.  Therefore  the  number  of  figures  in  the  root  of 
any  proposed  whole  number  may  be  found,  by  beginning 
at  the  right  and  separating  that  number  into  parts  or 
periods  of  two  figures  each.  The  left-hand  period  may 
contain  either  one  or  two  figures.  There  will  be  as  many 
figures  in  the  root  as  there  are  periods  in  the  power. 

Separating  105625  into  periods,  thus,  10^56^25,  we  see 
that  its  root  must  consist  of  three  figures,  viz.,  hundreds, 
tens,  and  units. 

Let  a  represent  the  hundreds,  h  the  tens,  and  c  the 
units  of  the  root ;  then  a-\-h-\-  c  will  represent  the  root. 

The  second  power  ofa-|-^~t"^  ^^  c^-\-^ah-\-y^ -\- 
2ac-|-26c-[-c2.  This  may  be  put  in  either  of  the  fol- 
lowing forms,  viz.,  a2  _|_  2  a  6  -[-  62  _|_  2  (a  -|-  6)  c  +  c^,  or 
«2  ^  (2  «  -I-  6)  6  +  [2  (a  -f  &)  +  c]  c.     The  first  of  these 


148  EXTRACTION    OF    THE  [<§>  27  ^ 

forms  shows  that  the  second  power  of  a  number  consisting 
of  hundreds,  tens,  and  units,  contains  the  second  power 
of  the  hundreds,  plus  twice  the  product  of  the  hundreds 
and  tens,  plus  twice  the  sum  of  the  hundreds  and  tens 
multiplied  by  the  units,  plus  the  second  power  of  the 
units. 

Let  us  now  proceed  to  extract  the  root  of  105625. 
Operation. 
10'56'25(325.     Root. 
9  =  a\ 

15'6(62        z=:2a  +  6. 
124  ~        =:(2a  +  6)6. 

322'5(645i=2(a  +  6)  +  c. 
3225  zz:[2(«4-6)  +  c]c. 

0. 

After  separating  the  number  into  periods,  we  first  seek 
the  second  power  of  the  hundreds  of  the  root,  which  must 
be  found  in  the  10,  (that  is,  100000) ;  the  greatest  second 
power  of  hundreds  in  this  part  is  9,  (90000),  the  root  of 
which  is  3,  (300). 

Putting  3  as  the  first  figure,  or  hundreds,  of  the  root, 
we  subtract  its  second  power  9  from  10 ;  to  the  right  of 
the  remainder  1  we  bring  down  the  second  period  SQ, 
and  have  156,  (15600),  which  we  regard  as  a  dividend. 

This  dividend  contains  2  a  6  -j-  ^^>  ^^  twice  the  product 
of  the  hundreds  and  tens,  plus  the  second  power  of  the 
tens;  it  also  contains  the  hundreds  which  are  produced 
by  multiplying  the  sum  of  the  hundreds  and  tens  by.  the 
units.    1 

We  now  wish  to  obtain  6,  the  tens  of  the  root ;  and,  for 
this  purpose,  we  divide  by  twice  the  hundreds,  correspond- 
ing to  2  a.  But  since  the  product  of  hundreds  by  tens 
has  three  zeros  at  the  right,  this  product  can  have  no  sig- 


<^27.J       SECOND  ROOTS  OF  NUMBERS.  149 

nificant  figure  below  the  fourth  place  from  the  right.  We 
therefore  reject  the  6,  separating  it  by  an  accent,  and 
divide  15,(15000),  by  6,(600),  twice  the  hundreds  of  the 
root. 

The  quotient  is  2,  (20),  which  we  put  as  the  second 
figure  or  tens  of  the  root ;  we  also  place  it  at  the  right  of 
the  divisor.  The  divisor  thus  increased  is  62,  (620),  and 
^oswers  to  2  a  -|-  6,  which  we  multiply  by  6  or  b,  and  have 
124,(12400),  answering  to  (2a-f-6)6. 

We  subtract  124  from  the  dividend  156,  and  have  for 
a  remainder  32,  to  the  right  of  which  we  annex  the  last 
period  25,  and  regard  the  result  3225  as  a  new  dividend. 
,  This  dividend  corresponds  to  '^  {a -\- h)  c  -\-  c^,  or 
[2  (a  -f"  ^)  "I"  ^]  ^  >  that  is,  it  contains  twice  the  sum  of  the 
hundreds  and  tens  multiplied  by  the  units,  plus  the  second 
power  of  the  units.  To  find  the  units  of  the  root,  there- 
fore, we  must  divide  by  twice  the  sum  of  the  hundreds 
and  tens,  that  is,  by  twice  the  whole  root,  so  far  as  it  has 
been  already  ascertained. 

But  since  hundreds  and  tens,  multiplied  by  units,  must 
have  one  zero  at  the  right,  this  product  can  have  no  sig- 
nificant figure  below  the  second  from  the  right ;  we  there- 
foije  reject  the  figure  5,  separating  it  by  an  accent.  Twice 
the  hundreds  and  tens  make  64,(640),  corresponding  to 
2  (a -|- 6),  which  is  contained  in  322    (3220)   five  times. 

We  put  5,  which  is  the  presumed  value  of  c,  as  the 
next  figure  of  the  root,  also  at  the  right  of  the  divisor. 
The  divisor  thus  increased  answers  to  2  (« -f-  6)  -|-  c, 
which  multiplied  by  5,  or  c,  gives  3225,  corresponding  to 
[2  («  +  ^)  +  c\  c  ;  this  subtracted  from  the  dividend 
leaves  no  remainder.     Hence  325  is  the  root  required. 

Of  the  two  formulae,  already  given,  of  the  second  power 
of«  +  6  +  c,  viz.,a2_[_2a6  4-2(a  +  6)c  +  6-2,  anda2_^ 
C2«  +  6)6+ [2(a4-6)  +  c]c,  the   former  shows,   that, 


150  EXTRACTION    OF    THE  [<§>  27. 

after  the  first  figure  has  been  found,  each  succeeding 
figure  is  to  be  sought  by  dividing  by  twice  the  whole  of 
the  root  pieviously  found;  and  the  latter  shows  that,  in 
each  case,,  the  quotient  is  to  be  placed  at  the  right  of  the 
divisor,  and  that  the  divisor,  thus  increased,  is  to  be  multi- 
plied by  the  quotient. 

Moreover,  from  the  rank  of  the  figures,  it  is  evident 
that  twice  the  root  already  found  can  produce  no  signifi- 
cant figure  below  the  second  from  the  right  in  each 
dividend. 

Formulae  might  be  given  for  the  second  power  of  four 
or  more  figures ;  but  from  what  has  been  already  shown, 
the  mode  of  proceeding,  in  all  cases,  will  be  sufficiently 
manifest. 

We  exhibit  below  the  process  of  extractmg  a  root  con- 
sisting of  five  figures. 

Operation. 

28  1 1  22  64  41  (  53021.     Root. 

25  ; 

311(103 
309 

22264  (  10602 

21204 


106041  ( 106041 
106041 

0. 

In  this  example  we  find  that  the  second  divisor  105  is 
not  contained  in  the  dividend  222,  the  right-hand  figure 
being  rejected  ;  this  shows  that  there  are  no  hundreds  in 
required  root.  In  such  a  case,  we  place  a  zero  in  the 
root,  also  at  the  right  of  the  divisor,  and  annex  the  sue 
ceeding  period  to  form  a  dividend. 


§27.]  SECOND    ROOTS    OF    NUMBERS.  151 

Art.  85.  From  the  foregoing  examples  and  exp\ana- 
tions  we  deduce  the  following 

RULE    FOR    EXTRACTING   THE    SECOND    ROOTS    OF    NUMBERS. 

1.  JBegin  at  the  right ,  and,  hy  means  of  accents  or 
other  marks,  separate  the  number  into  periods  of  two 
figures  each.  The  left-hand  period  may  contain  one  or 
two  figures. 

2.  Find  the  greatest  second  power  in  the  left-hand 
period,  place  its  root  at  the  right  of  the  proposed  number, 
separating  it  by  a  line,  and  subtract  the  second  power 
from  the  left-hand  period. 

3.  To  the  right  of  the  remainder  bring-  down  the  next 
period  to  form  a  dividend.  Double  the  root  already 
found  for  a  divisor.  Seek  how  many  times  the  divisor  is 
contained  in  the  dividend,  the  right-hand  figure  being  re- 
jected. Place  the  quotient  in  the  root,  at  the  right  of  the 
figure  previously  found,  and  also  at  the  right  of  the 
divisor.  Multiply  the  divisor,  thus  increased,  by  the  last 
figure  of  the  root,  and  subtract  the  product  from  the  whole 
dividend, 

4.  Repeat  the  process  in  part  third  of  the  rule,  until 
all  the  periods  have  been  brought  down. 

Remark  \st.  If  the  dividend  will  not  contain  the  di- 
visor, the  right-hand  figure  of  the  former  being  rejected, 
place  a  zero  in  the  root,  also  at  the  right  of  the  divisor, 
and  bring  down  the  next  period. 

Remark  2d.  We  may  observe,  that,  if  the  last  figure 
of  the  preceding  divisor  be  doubled,  the  root  will  be 
doubled ;  for  that  divisor  contains  twice  the  whole  root 
found,  with  the  exception  of  the  figure  last  obtained. 

Remark  Sd.  To  find  the  root  of  a  product,  as  will  be 
shown  hereafter,  we  take  the  root  of  each  factor  and  mul- 
tiply these  roots  together.  Thus,  the  root  of  81 .  225  ia 
9    15=135. 


I5"2    EXTRACTION    OF    THE    SECOND    ROOTS,    &C.     [4>  27 

Extract  the  roots  of  the  subjoined  numbers. 

1.  676.  8.  266256.  15.  121.144. 

2.  1369.  9.  289444.  16.  81.256. 

3.  2304.  10.  628849.  17.  400.361. 

4.  10816.  11.  12759184  18.  16.25.9. 

5.  22201.  12.  8590761.  19.  4.64.144. 

6.  26569.  13.  9616201.  20.  676.441. 

7.  66049.  14.  81036004. 

Art.  8G.  Comparatively  few  numbers  are  exact 
second  powers,  and  the  roots  of  such  as  are  not  perfect 
powers  cannot  be  found  exactly,  either  in  whole  numbers 
or  fractions.  Thus,  the  second  root  of  5  is  between  2 
and  3 ;  but  no  numb^  can  be  obtained,  which,  multiplied 
by  itself,  will  produce  exactly  5.  Such  a  root  may,  how- 
3ver,  be  approximated  to  any  degree  of  accuracy. 

All  whole  numbers  and  all  definite  fractions  are  called 
commensurable  or  rational,  because  they  have  a  common 
measure  with  unity,  or  their  ratio  to  unity  can  be  exactly 
obtained.  But  the  root  of  a  number  which  is  not  a 
perfect  power,  is  called  incommensurable  or  irrational, 
because  it  has  no  common  measure  with  unity,  or  its 
ratio  to  unity  cannot  be  exactly  found.  Such  roots,  oi 
rather  expressions  of  them,  are  ca/led  also  surds. 

The  second  root  of  a  quantity  is  denoted  either  by  the 
exponent  ^,  or  by  the  sign  ^,  called  the  radical  sign 
Thus,  4*^  or  ^4  =  2,  and  3^  or  a/  3  means  the  second 
root  of  3. 

The  second  root  of  a  negative  quantity  is  called  imagi^ 
nary,  because  no  quantity,  either  positive  or  negative, 
can,  when  multiplied  by  itself,  produce  a  negative  quan- 
tity. Thus,  ( — 4)2  or  ^  — 4  is  an  imaginary  quantity. 
An  imaginary  result  to  a  problem  generally  indicatf.s 
absurdity  in  the  conditions  of  that  problem. 


^28  1     SECOND  ROOTS  OF  FRACTIONS,  &C.      153 


SECTION   XXVIII. 

•ECOND  ROOTS  OF  FRACTIONS,  AND  EXTRACTION  OF 
SECOND  ROOTS  BY  APPROXIMATION. 

Art.  87.  A  fraction  is  raised  to  the  second  power  by 
raising  both  numerator  and  denominator  to  that  power, 
this  beincr  equivalent  to  multiplying  the  fraction  by  itself. 

Thus,  (IV^l.A^A;  and(-V=^.:^  =  ™'. 

Consequently,  the  second  root  of  a  fraction  is  found  by 
extracting  the  root  of  both  numerator  and  denominator. 

Thus,  the  second  root  of  —  is  — ;  that  of  —  is  — . 
'  16         4  '  62         6 


■fi" 


-; 


Extract  the  second  roots  of  the  following  fraction^  / 

•      2.    14.         .  6.   ifff. 

3.    Iff.     #V       .  7.    Iff. 

vj  If  i 

Art.  88.  But  if  either  the  numerator  or  denominator 
is  not  an  exact  second  power,  we  can  find  only  an  approxi- 
mate root  of  the  fraction.  Thus,  the  root  of  -ff  is  be- 
tween f  and  I,  or  1,  the  latter  being  nearer  the  true  root 
than  the  former. 

We  can  always  make  the  denominator  of  a  fraction  a 
perfect  second  power,  by  multiplying  both  numerator  and 
denominator  by  the  denominator.  This  does  not  change 
the  value,  but  only  the  form  of  the  fraction.    For  example 

—  =  -^  =  — ,  the  approximate  root  of  which  is 1— 

8        8.8       64  ^*^  8    ' 

Remark.     The  sign   -f-,  placed  after   an  approximate 


154         SECOND  ROOTS  OF  FRACTIONS,       [^ '^S 

root,  denotes  that  it  is  less,  and  the  sign  — ,  that  it  is 
greater,  than  the  true  root. 

If  a  greater  degree  of  accuracy  is  required,  we  may, 
after  preparing  the  fraction  as  above,  multiply  both  nu- 
merator and  denominator  of  the  result  by  any  second 
power,  and  then  extract  the  root.  Thus,  to  find  the  root 
of  ^ ;  after  changing  it  to  f  f ,  we  may  multiply  the  nu- 
merator and  denominator  of  ff  by  225,  the  second  power 
14175 

of  15 ;  this  gives  — — -  ,  the  approximate  root  of  which 

.       119         119    , 

IS  =z U. 

9.15        135    ' 

Art.  89.  The  roots  of  whole  numbers,  which  are  not 
exact  second  powers,  may  be  approximated  in  a  similar 
manner.  For  example,  to  find  the  root  of  3,  accurate  to 
within  YTf  we  convert  it  into  a  fraction  having  the  second 
power  of  15  for  its  denominator.  Thus,  3  ==  fjf ,  the 
root  of  which  is  f  |  — . 

But  the  most  convenient  number  for  a  denominator  is 
the  second  power  of  10,  100,  or  1000,  &/C. ;  that  is,  we 
convert  the  number  into  lOOths,  lOOOOths,  or  lOOOOOOths, 
&;C.,  and  the  root  will  be  in  decimals. 

'T'l,,,^      Q 300  30000  3000000.     fViof     ;«     Q *i't\(\ 

—  30000  =  3000000.  The  approximate  root  of  the  first. 
IS  -fj  +  z=  1-7  + ;  that  of  the  second  is  j-i^  +  =  1*73  + ; 
that  of  the  third,  j^iU  +  =  1*732  +. 

It  is  evident  that  there  must  be  twice  as  many  decimals 
in  the  power  as  we  wish  to  find  in  the  root ;  for  the  second 
power  of  lOths  produces  lOOths,  the  second  power  of 
lOOths  produces  lOOOOths,  &/C.  Hence  two  zeros  must 
be  annexed  to  the  number  for  each  additional  decimal  in 
the  root.  Nor  need  the  zeros  be  all  written  at  once,  but 
we  may  annex  two  zeros  to  each  remainder,  in  the  same 
manner  as  we  bring  down  successive  periods. 


§28.J  AND    APPROXIMATE    SECOND    ROOTS.  155 

Lei  us,  in  this  way,  extract  the  second  root  of  5. 
Operation, 
5(2-236+.     Root. 
4 

*  10^0(42 

84_ 

160^0  (443 
1329 

2710'0  (4466 
26796 
304. 

The  operation  may  be  continued  to  any  desirable 
extent. 

When  the  given  number  contains  decimals,  the  process 
of  finding  the  root  is  the  same ;  and  any  fraction  may  be 
changed  to  decimals  and  the  root  be  found  in  the  same 
way,  care  being  taken,  in  both  cases,  to  make  the  number 
of  decimals  even,  and  to  point  off  half  as  many  figures 
for  decimals  in  the  root,  as  there  are  in  the  power,  includ- 
ing the  zeros  annexed. 

In  separating  a  number  containing  decimals  into  pe- 
riods, it  is  best  to  begin  at  the  decimal  point,  and  separate 
the  decimals  by  proceeding  towards  the  right,  and  the 
integral  numbers  by  proceeding  towards  the  lefl. 

Let  the  roots  of  the  following  numbers  be  found  in 
decimals,  each  root  containing  three  decimal  figures. 


1. 

7. 

3. 

24. 

3. 

23. 

4. 

25-72. 

5. 

31-2. 

6. 

1- 

-7-  7. 

^2- 

-8. 

3|. 

9. 

105|. 

10. 

lOf' 

t^ 

156     PURE  EQUATIONS  OF  THE  SECOND  DEGREE.    [<§  29. 


/      SECTION   XXIX. 

QUESTIONS    PRODUCING    PURE    EQUATION*   OF    THE    SECOND 
DEGREE 

Art.  90.  1.  Two  numbers  are  to  each  other  as  2  to 
3,  and  their  product  is  96.     Required  the  numbers    /ij^    i 

2.  What  number  is  tTiat  whose  ^  part  multiplied  by  its  ^ 
J  part  produces  48  ?     cc^  ^ 

3.  The  length  of  a  house  is  to  its  breadth  as  10  to  9, 
and  it  covers  1440  square  feet  of  land.  Required  the 
length  and  breadth.  ]^  ^ 

4.  What  number  is  that  to  which  if  5  be  added,  and 
from  which  if  5  be  subtracted,  the  product  of  the  sum 
and  difference  shall  be  39?         ^ 

5.  A  man  bought  a  farm,  giving  ^  as  many  dollars  per 
acre  as  there  were  acres  in  the  farm,  and  the  whole 
amounted  to  $2000.  Required  the  number  of  acres  and 
the  price  per  acre.  CO 

6.  Two  numbers  are  to  each  other  as  8  to  5,  and  the 
difference  of  their  second  powers  is  156.  Required  these 
numbers.  ' 

7.  A  gentleman  has  a  rectangular  piece  of  land  25  rods 
long  and  9  rods  wide,  which  he  exchanges  for  a  square 
piece  of  the  same  area.  Required  the  length  of  one  side 
of  the  square.  ' 

/>&r--The  sides  of  two  square  floors  are  to  each  other  as 
7  to  8,  and  it  requires  15  square  yards  more  of  carpeting 
to  cover  the  larger,  than   it  does  to  cover  the  smaller.     , 
Required  the  length  of  one  side  of  each  floor.     /"     >       /' 

9.  A  farmer  bought  two  equal  pieces  of  land,  giving 
for  the  whole  $1800.     For  one  he  gave  $10  less,  and  for 


VL  "  r  /  —'  ^ 

V  -^  C'  ^ 

<5>30.]  AFFECTED    EQUATIONS. 

the  other  $10  more,  per  acre,  than  there  were  acres  ia 
each  piece.     Required  the  number  of  acres  in  each.   ^L) ' 

10.  The  product  of  two  numbers  is  500,  and  the 
greater  divided  by  the  less  gives  5.  What  are  the  num- 
bers ?  h  5 

11.  An  acre  contains  160  square  rods.  Required  the 
length  of  one  side  of  a  square  containing  an  acre. 

12.  What  is  the  length  of  a  square  piece  of  land  in 
which  there  are  5  acres  ? 

13.  A  cistern  having  a  square  bottom,  and  being  4  feet 
deep,  contains  600  gallons.  Required  the  length  of  one 
side  of  the  bottom,  a  gallon  wine  measure  i)eing  231 
cubic  inches.       / 


SECTION  XXX.  - 

AFPpCTED    EQUATIONS    OF    THE    SECOND    DEGREE. 

Art.  91.  Equations  of  the  second  degree  which  we 
have  hitherto  considered,  contained  the  second  power,  but 
no  other  power,  of  the  unknown  quantity.  But  an  equa- 
tion of  the  second  degree,  in  its  most  general  sense,  con- 
tains three  kinds  of  terms,  viz.,  one  in  which  there  are  two 
unknown  factors ;  another  in  which  there  is  out  one  un- 
known factor ;  and  a  third  composed  wholly  of  known 
quantities. 

Equations  of  this  description,  when  they  contain  only 
one  unknown  quantity,  are  called  affected  equations  of  the 
second  degree,  or  affected  quadratic  equations. 

1.  The  length  of  a  rectangular  piece  of  land  exceeds 
its  breadth  by  6  rods,  and  the  piece  contain?  112  square 
rods.     Required  the  dimensions. 


158  AFFECTED    EQUATIONS  [^30. 

Let  X  (rods)  =  the  breadth  ;  then 
a;  -[-  6  1=1:  the  length,  and 
x^  -|-  6  2:  z=  the  area.     Hence, 
a;2-f-6a:z=112. 
In  order  to  solve  this  equation,  we  compare  the  first 
member  of  it  with  the  second  power  of  a;  -j-  «,  which  is 
x^-\-2ax-\'  a^.     We  perceive  that  there  are  two  terms  in 
each,  which  correspond  to  each  other,  and  we  shall  ex- 
press this  correspondency  by  the  sign  of  equality.     Thus, 
x^=ix\ 
2ax=:6x.     Hence,  by  division, 
^       2a  =  6 
a  =  S.'. 
a^  =  9. 
Since  9  corresponds  to  a^,  if  we  add  9  to  each  member 
of  x^-^Gx^z  112,  we  have 

2:2  4_6a:  +  9  =  121, 
the  first  member -€)f  which  corresponds  to  x^-\-2ax-\-a^, 
and  is  a  perfect  second  power. 

We  now  extract  the  root  of  each  member.  The  root 
of  the  first  member  is  a^  -|-  3,  for  {x  -[-  3)  (x  -f-  3)  :=  a:^  -f- 
6  a;  -f-  9,  and  this  root  corresponds  to  a;  -f-  a ;  the  root  of 
the  second  member  is  11.     Hence, 

a; -|- 3  =  11, .'.  a;  =  8  rods,  breadth, 
a:  -[-  6  =:  14  rods,  length. 
We  remark  that  every  positive  quantity  has  two  second 
roots,  one  positive,  and  the  other  negative.  For  the 
second  power  of  -\-  «,  and  the  second  power  of  —  a,  are 
each  -^-a^.  Therefore,  the  second  root  of  a^  is  either 
-}-«,  or  — a.  In  like  manner,  the  second  root  of  121  is 
either  -j-  11,  or  —  11. 

Now,  since,  in  an  equation  similar  to  a;-f-3=  II,  the 
value  of  X  is  not  ascertained,  until  the  known  term  has 
been  transposed  from  the  first  to  the  second  member,  it 
may  happen  that  the  negative  as  well  as  the  positive  value 


<5>30.]  OF    THE    SECOND    DEGREE.  159 

of  the  second  member  will  answer  the  conditions  of  the 
question. 

In  all  cases,   therefore,  of  affected    equations   of  the 
second  degree,  we  prefix  to  the  root  of  the  second  mem- 
ber the  double  sign  zb.  which  is  read  ^^ plus  or  minusJ' 
Giving  this  sign  to  11,  we  have 

a;4-3  =  ±ll,  and 

X=:±ll— 3. 

Calling  11  positive,  we  have  x=iS ; 

calling  it  negative,  we  have  xz=z  — 14. 

The  former  value  only  of  x  will  satisfy  the  conditions 

of  the  question.     The  latter  value  will,  however,  satisfy 

the  original  equation;  for, 

(_  14)2  _^  6  (— 14)  :zz  196  —  84  =  1 12. 
2.   What  number  is  that  whose  second  power  increased 
by  20  is  equal  to  12  times  the  number  itself? 
Let  X  :=;  the  number.     Then, 
^2  -j-  20  zz:  12  X,     By  transposition, 
a:2— 12a:=:  — 20. 
In  this  equation  12  x  is  negative,  and  in  order  to  render 
the  first  member  a  complete  second  power,  we  compare  it 
with  the  second  power  of  x  —  a,  which  is  x^  —  ^ax-\-a^. 
This  comparison  gives 

a;2  =  a;2, 
—  2ax=.  —  12a;;  .•. 
_2a  =  — IS, 
_«  =  — 6, 

Adding  36  to  each  member  of  x^  — 12  a;  =  —  20,  we 
have 

a:2— 12  X  +  36  =  — 20  +  36  =  16. 

The  first  member  now  corresponds  to  x^  —  2  a  x  -j-  a^, 
and  is  a  perfect  second  power.  The  next  step  is  to  ex- 
tract the  root  of  each  member.     The  root  of  the  first 


160  AFFECTED    EQUATIONS  [^30. 

member  is  x  —  6,  for  (z  — 6)  (x— 6)  =:x2— 12x  +  36 
This  root  corresponds  to  x  —  a;  the  root  of  the  second 
member  is  zh  4.     Hence, 

X  —  6 1=  zb  4.     Transposing, 

x  =  6±4.,r. 

XzzzlO^OY  x  =  2. 
Both  values  of  x  have  the  sign  -|-,  and,  therefore,  both 
answer  the  conditions  of  the  question.  Indeed,  (10)2-|- 
20  =  12 .  10  z=  120,  and  (2)2  -[-  20  =  ]  2  . 2  =  24.  Hence 
we  see  the  propriety  of  giving  the  double  sign  to  the  root 
of  the  second  member. 

Art.  92.  Any  affected  equation  of  the  second  degree 
may  be  reduced  to  the  form  of  x^  -^p  x=:q,  in  which  p 
and  q  may  stand  for  any  known  quantities,  positive  or 
negative. 

An  affected  equation  may,  manifestly,  be  reduced  to 
this  form  in  the  following  manner,  x  being  the  unknown 
quantity. 

1.  Clear  the  equation  of  fractions,  if  necessary  ;  trans^ 
pose  all  the  terms  containing  x^  and  x  into  the  first  mem-' 
ber,  and  the  known  terms  into  the  second;  reduce  the  terms 
which  contain  x^  to  one  term,  and  those  which  contain  x 
to  another ;  also,  reduce  the  known  quantities  in  the  second 
member  to  a  single  term, 

2.  If  the  term  containing  x^  is  not  positive,  make  it  so 
by  changing  the  signs  of  all  the  terms, 

3.  If  the  coefficient  of  x^  is  not  1,  divide  all  the  terms 
by  that  coefficient. 

1.  Let  it  be  required  to  solve  the  following  equation ; 
nz. 

-^+1=3. 

X-\-l      '        X 

C  earing  the  equation  of  fractions  by  multiplication 


*§>30.]  OF    THE    SECOND    DEGREE.  IP^ 

62:-f-2x  +  2i=:3a:2-f3x.     Transpose, 
^^Sx^  —  '^x-\'6x'}'2x=:—2;  reduce 

*ix^-\-5x=:  —  2;  change  the  signs, 
Sx^  —  5xi=z2;  divide  by  3, 
x^—^x  =  %. 
The  equation  is  now  reduced  to  the  form  of  x^-^-px 
=:  q,  and  we  have  p^=^  —  f ,  and  g^  =  f . 

Comparing  the  first  member  of  x^  —  j^  x  =r.  §  with  x^  — 
2  a  X  -|-  «^j  we  have 

—  2aa;z=i  —  fx; 

—  2a  =  — 5; 

Adding  f |^  to  each  member  of  x^  —  ^x-=:%.  we  have 
reducing  the  second  member  to  a  single  fraction, 

^'-f^  +  M  =  lf- 

Extracting  the  root  of  each  member,  we  have  for  the 
root  of  the  first  x  —  ^,   for   (x  —  |)  {x  —  ^)z=lx^  —  ^x 
+  f  I,  and  for  the  root  of  the  second  ^iz  i-     Hence, 
X  —  1^  z=  ^3 1^.     Transpose  —  f , 
x=%±l,.'. 

x  =  :^-  =  ^,  or  x=:  — fzzi  — f 
2.    Since  every  affected  equation  of  the  second  degree 
may  be  reduced  to  the  form  of  x^  -\-p  x-=:.q,  let  us  solve 
this  general  equation.     Comparing  the  first  member  of  it 
with  x^-\-2aX'^a^,  we  have 

X^  =  X^', 

2  axzzip  x; 
^a=p) 
a  =  ip; 
a^  =  ip^. 
Adding  ^p^  to  each  member  of  x^  -\-p  x=:q^  we  have 
11 


162  AFFECTED    EQUATIONS  [<§>  30 

Extracting  the  root  of  the  first  member,  and  expressing 
that  of  the  second,  (Art.  86), 

^  +  iP  =  ±{fI  +  iP^)^'^  hence, 
x  =  —  ipdz{q  +  ip^)^- 

Remark,  It  is  evident  that  the  root  of  the  second 
member  must  be  indicated  merely,  until  definite  values  are 
assigned  to  p  and  q.  The  second  root  of  a  quantity  is 
properly  indicated  by  the  exponent  ^.  For  the  product 
of  the  second  root  of  a  quantity  by  itself,  must  pro- 
duce that  quantity ;  and,  according  to  the  rule  for  the 
exponents  in  multiplication  (Art.  30),  a^,a^:=: a}  or  a; 
therefore,  a^  is  the  second  root  of  a.  For  a  similar 
reason,  (q-^^p^)^  represents  the  second  root  of  g'  +  ip^ 
The  radical  sign  may  be  used  by  those  who  piorer  it 
Thus,  Y  q^iP^  likevi^ise  denotes  the  root  of  q-\-ip'^- 

Art.  93.  From  the  solution  of  the  preceding  genera 
equation,  we  derive  a 

RULE   FOR    SOLVING  AFFECTED    EQUATIONS    OF  THE    SECOND 
DEGREE 

1.  Reduce  the  equation  to  the  form  of  x^  -^ p  x  z=z  q, 

2.  Make  the  Jirst  member  a  perfrct  second  power,  by 
adding  to  both  members  the  second  power  of  half  the 
coefficient  of  x  {or  of  the  first  power  of  the  unknown 
quantity). 

3.  Extract  the  root  of  each  member.  The  root  of  the 
first  member  will  consist  of  two  terms,  the  first  of  which 
is  X,  or  the  unknown  quantity,  and  the  second  is  half  the 
coefficient  previously  found,  having  the  same  sign  as  that 
coefficient,  and  the  root  of  the  second  member  must  have  the 
double  sign  ztz* 


^30.]  OF    THE    SECOND    DEGREE.  163 

4.  Transpose  the  known  term  from  the  first  to  the 
second  member  and  r educe y  and  the  value  of  x  will  he 
found. 

Since  p  and  q  may  be  either  positive  or  negative,  the 
general  equation  admits  of  four  forms,  differing  only  with 
regard  to  the  signs  of  p  and  q,  that  of  ^p^  being  always 
positive.  These  forms,  and  the  consequent  values  of  x, 
are  as  follows. 

(1)  x^-\'px  =  q\  whence,  a:=:  —  i^i'zb  (g'  +  ii^^)*- 

(2)  x^ — px=zq;  whence,  a:  =  +  i"J^=t  (S'  +  ii^^)^- 

(3)  x^'-\-px=z  —  q;  whence,  a;  =  —  iPzhUp^  —  q)^- 

(4)  x^ — px=z  —  qi  whence,  X  z= -f"  iP  ±  (iP^  —  q)^' 

Art.  04.     1.    Solve  the  following  equation;  viz. 

X       I    a;+l 13 

x-\-l~^     X  6  * 

Clear  the  equation  of  fractions,  and  reduce, 

12  x2  +  12  x  +  6  ~  13  2:2  -|L  13  a; ;  transpose, 
122:2-fl2a;  — 13x2— 13a;z=  — 6;  reduce, 
—  x^  —  x=i  —  6 ;  change  the  signs, 
x2  +  X  m  6. 
The  equation  is  now  in  the  form  of  x^ -\- p  x  =z  q.     The 
coefficient  of  x  being  1,  we  are  to  add  to  each  member 
the  second  power  of  J,  which  is  ^.     We  then  have 
x^-\'X-\-i  =  6-{-i  =  \^-,     Extracting  the  root, 
x  +  i  =  ±i.'. 
x  =  —  i±^  =  2,oT—S. 
2.  What  number  is  that  whose  second  nower,  increased 
by  3^,  is  equal  to  7  times  that  number  ? 

Let  X  =  the  number.     Then, 
x2  -f.  3^  =:  7  x.     By  transposition, 
x^  —  7x=:  —  Si. 
In  thi?  equation  it  is  not  necessary  to  remove  the  de» 


164  AFFECTED    EQUATIONS  [<§>  30. 

nominator  4,  the  equation  being  now  in  the  general  form. 
Half  of  the  coefficient  of  z  is  — f,  the  second  power  of 
which,  added  to  both  members,  gives 
x^  —  7x  +  ^=i  —  Si  +  ^^-=i -^6 .     Taking  the  root, 

3.  Separate  the  number  10  into  two  parts,  such  that 
their  product  shall  be  30. 

Let  X  be  one  part ;  then  10  —  x  will  be  the  other.    Hence, 

10  X  —  x^  =  30.     Changing  the  signs, 

x^  —  10  a:  =  —  30.     Completing  the  second  power, 

a;2  _  10  x  +  25 1=  —  30  +  25  z=z  —  5.     Taking  the  root, 

x  —  5=::±:^—5.'.x=i6±^—5. 

Since  the  second  root  of  a  negative  quantity  is  imagi 

nary,  this  problem   is  impossible  ;    indeed,   the  greatest 

quantity  that  can  result  from  the  product  of  the  two  parts 

pf  a  number,  is  ^  of  the  second  power  of  that  number. 

4.  The  length  of  a  rectangular  field  exceeds  its  breadth 
by  8  rods,  and  the  field  contains  180  square  rods.  Re- 
quired the  dimensions.         IQ 

5.  A  draper  sold  a  piece  of  cloth  at  $4  less  per  yard 
than  there  were  yards  in  the  piece,  and  the  price  of  the 
whole  was  $60.  Required  the  number  of  yards  and  the 
price  per  yard.  /     ~' 

6.  There  are  two  square  rooms,  the  side  of  one  being 
1  yard  longer  than  the  side  of  the  other.  To  carpet  both 
rooms  requires  85  yards  of  carpeting.  How  long  is  one 
side  of  each  room  ? 

7.  What  nuiUDer  is  such,  that  if  you  subtract  it  fi-om 
10,  and  multiply  the  remainder  by  the  number  itself,  the 
remainder  shall  be  21 1  / 

8.  A  farmer  bought  a  Certain  number  of  sheep  for 
$80 ;  but  if  he  had  bought  5  more  for  the  same  sum,  the 


«5>30.]  OF    THE    SECOND    DEGREE.  165 

price  of  one  sheep  would  have  been  f  of  a  dollar  less. 
How  many  sheep  did  he  purchase? 

9.  What  two  numbers  are  those  whose  sum  is  29  and 
whose  product  is  100  ?         ^  /*  -      ^ 

10.  A  man  built  two  pieces  of  wall,  one  of  which  was 
2  rods  longer  than  the  other.     He  received  as  many  shil- 
lings per  rod  for  each  as  there  were  rods  in  the  longer ^>5^3^^^^_ 
and  the  whole  amounted  to  $44.     Required  the  length  of    "" 
each  piece. 

1 1.  What  number  is  that  to  which  if  3  times  its  second 
root  be  added,  the  sum  will  be  40  ? 

Let  x^  =  the  number  ;  then,  x^-^Sx  =  40. 

12.  There  are  two  numbers,  such,  that  if  the  less  be 
subtracted  from  3  times  the  greater,  the  remainder  will 
be  35 ;  and  if  4  times  the  greater  be  divided  by  3  times 
the  less,  plus  1,  the  quotient  will  be  the  less  number.  Re- 
quired the  numbers. 

The  solution  of  this  question  requires  two  unknown 
quantities. 

Let  X  =z  the  greater,  and  y  =z  the  less.     Then, 

(1)  3a;_y_35.^ 

(2)  _l^z=y.       ( 
Remove  the  denominator  in  2d, 

(3)  ^X=zS7J^+7/.'. 

From  the  1st  we  have 

(5)   ^  =  ^. 

Making  an  equation  with  the  values  of  x  in  the  4th 
and  5th, 

3y-^  +  y__35-f-y 
4  3 

Solving  this  equation,  we  have,  successively, 


166  AFFFrXED    EQ,UATIONS.  [^30. 

^  9y2_yz=140; 

y— tV  =  ±H; 

y  — T-VdzH  =  4,  or— 3f. 
Taking  4  as  the  value  of  y,  and  substituting^  it  in  5th. 

X  zz: =z  Id. 

13.  Separate  100  into  two  such  parts,  that  the  sum  of 
their  second  roots  shall  be  14.         "  k^  ^ 

Let  x^  and  y^  be  the  parts. 

14.  The  sum  of  two  numbers  is  24,  and  the  sum  of^   ,  ^ 
their  second  powers  is  306.     Required  the  numbers,  /y  *^A/ 

15.  The  greater  of  two  numbers  divided  by  the  second     T 
power  of  the  less,  gives  2  for  a  quotient,  and  ^  of  the  dif-         f, 
ference  of  the  numbers  i#5.     What  are  these  numbers  1   ^^' 

16.  The  less  of  two  numbers  added  to  twice  the 
greater,  makes  22,  and  half  of  their  product,  increased 
by  the  second  power  of  the  less,  makes  60.  Required 
the  numbers.  y-    /  V^  '      ///    ;*        -      V 

17.  Two  numbers  are  such,  that  twice  the  second 
power  of  the  greater  and  3  times  the  second  power  of  the 
less  make  147,  and  3  times  the  greater,  diminished  by 
twice  the  less,  make  8.     What  are  the  numbers  ? 

\ 


^31.]  EXTRACTION    OF    THIRD    ROOTS.  167 


SECTION   XXXI. 

EXTRACTION    OF    THE    THIRD    HOOTS    OF    NUMBERS. 

Art.  95.  The  product  of  a  number  multiplied  twice 
by  itself,  is  called  the  third  power  or  cube  of  that  number. 
Thus,  27  =  3  .  3  .  3  is  the  third  power  of  3,  and  a^  = 
a.  a,  a  is  the  third  power  of  a. 

The  first  power  of  a  quantity,  with  reference  to  the 
third,  is  called  the  third  root  or  cube  root.  For  example, 
3  is  the  third  root  of  27,  and  a  is  the  third  root  of  a^ ; 
and  the  process  of  deducing  the  first  power  from  the  third, 
is  called  extracting  the  third  or  cube  root, 

rhe  third  powers  of  the  first  nine  integral  numbers  are 
as  follows. 

Roots.     1, 2,    3,    4,      5,      6,      7,      8,      9.  ) 
Third  powers.     1,  8,  27,  64,  125,  216,  343,  512,  729.  ) 

The  second  line  contains  the  third  powers,  of  which 
the  corresponding  numbers  in  the  first  line  are  the  third 
roots.  We  perceive,  moreover,  that,  of  integral  numbers 
consisting  of  one,  two,  or  three  figures,  there  are  only 
nine  which  are  perfect  third  powers.  The  roots  of  such 
numbers  as  are  not  exact  third  powers  cannot  be  obtained 
exactly,  although  they  may,  as  we  shall  see  hereafter,  be 
approximated  to  any  degree  of  accuracy.  Thus,  the  third 
root  of  29  is  between  3  and  4,  3  being  nearer  than  4  to 
the  true  root. 

The  third  power  of  a  number  having  zeros  on  the 
right,  contains  three  times  as  many  zeros  as  that  num- 
ber. Thus,  (10)3  --  1000,  (100)3  ~  1000000,  (1000)3  ^ 
1000000000,  &c.  Consequently,  the  third  power  of  a 
number  between  10  and  100,  that  is,  of  a  number  con- 


168  EXTRACTION    OF    THE  [^31 

taining  two  figures,  must  be  greater  than  1000,  and  less 
than  1000000 ;  in  other  words,  it  cannot  consist  of  less 
than  four  nor  of  more  than  six  figures.  Also,  the  third 
power  of  a  number  between  100  and  1000,  that  is,  of  a 
number  consisting  of  three  figures,  must  be  greater  than 
1000000  and  less  than  1000000000;  in  other  words,  it 
cannot  consist  of  less  than  seven  nor  more  than  nine 
figures.  In  like  manner,  when  the  root  contains  four 
figures,  the  power  must  contain  either  ten,  eleven,  or 
twelve  figures. 

Hence,  we  may  readily  ascertain  the  number  of  figures 
in  the  root  of  any  number,  by  commencing  at  the  right 
and  separating  the  number  into  periods  of  three  figures 
each.  The  left-hand  period  may  consist  of  one,  two,  or 
three  figures.  There  will  be  as  many  figures  in  the  root 
as  there  are  periods.  This  separation  may  be  denoted  by 
accents  or  other  marks. 

If  a  number  is  a  perfect  third  power,  and  contains  no 
more  than  three  figures,  its  root  may  be  found  by  inspec- 
tion or  trial.  When  the  numbejr  consists  of  more  than 
three  figures,  its  root  must  also  be  found  by  trial,  but  a 
rule  may  be  obtained  which  will  greatly  facilitate  th^ 
operation. 

Art.  96.  We  now  proceed  to  investigate  the  mocfe 
of  extracting  the  root  of  a  number  consisting  of  more 
than  three  figures,  that  of  17576,  for  example,  which  is 
the  third  power  of  26. 

Let  a  represent  the  tens,  and  b  the  units  of  the  root ; 
then  a  +  6  zz:  20  +  6  =  26.  The  third  power  of  a  +  6  is 
a3  -|-  3  a2  6  -f  3  a  62  -j-  63.  Putting  20  instead  of  a,  and  6 
instead  of  6,  we  have 

a  3  — (20)3=:  8000, 
3a36:z=3.(20)2.6-=7200, 


^31.]  THIRD    ROOTS    OF    NUMBERS.  169 

3a62~3. 20. 62  =  2160, 
63—      63       =216. 
Therefore,  a^-\-Sa^b-{-Sab^  +  b^  =  8000  +  7200  + 
2160  +  216  =  17576.     Hence, 

The  third  power  of  a  number  consisting  of  tens  and 
units,  contains  the  third  power  of  the  tens,  plus  three  times 
the  second  power  of  the  tens  into  the  units,  plus  three  times 
the  tens  into  the  second  power  of  the  units,  plus  the  third 
power  of  the  units. 

Now,  supposing  the  root  of  17576  unknown,  let  us 
trace  the  process  of  finding  it. 

Operation. 
17^576(26.     Root. 
8  =a\ 

9576(^2  =Sa^, 
(26)3=17576 

0. 

Separating  the  given  number  into  periods,  we  perceive 
that  the  root  must  consist  of  two  figures,  tens  and  units. 
As  the  third  power  of  tens  has  three  zeros  at  the  right,  it 
can  have  no  significant  figure  below  tnowsands.  The 
third  power  of  the  tens  sought,  must  therefore  be  found  in 
the  17,(17000).  The  third  power  of  2  (tens)  or  20  is  8, 
(8000),  and  the  third  power  of  3  (tens)  or  30  is  27, 
(27000).  Hence,  the  greatest  third  power  of  tens  con- 
tained in  17,(17000)  is  8,(8000),  the  root  of  which  is 
2  (tens)  or  20.  We  put  2  (tens)  in  the  root,  at  the  right 
of  the  given  number,  and  subtract  its  third  power  8, 
(8000),  from  17576.     The  remainder  is  9576. 

As  we  have  subtracted  the  part  corresponding  to  a^ 
the  remainder  corresponds  to  S  a^  b -{- S  a  b^ -{- b^.  Our 
next  step  is  to  find  the  units  of  the  root,  corresponding  to  b 


170  EXTRACTION    OF    THE  ['^sSl. 

If  our  remainder  contained  Sa^b  only,  or  exactly  three 
times  the  second  power  of  the  tens  into  the  units,  we 
should  evidently  find  b  or  the  units,  by  dividing  this  re- 
mainder by  3  a^,  that  is,  by  three  times  the  second  power 
of  the  tens  already  found ;  for  Sa^b  divided  by  3  a^  gives  b. 
Nevertheless,  if  we  divide  by  three  times  the  second 
power  of  the  tens,  neglecting  the  remainder,  we  shall  find 
the  units  exactly,  or  a  number  a  little  greater  than  the 
units. 

Three  times  the  second  power  of  2  (tens)  is  12,(1200), 
which  we  place,  as  a  divisor,  at  the  right  of  9576,  con- 
sidered as  a  dividend.  The  number  1200  is  contained  7 
times  in  9576,  or,  what  is  the  same  thing,  12  is  contained 
7  times  in  95.  We  put  7  in  the  root,  at  the  right  of  the 
2,  and  raise  27  to  the  third  power.  But  the  third  power 
of  27  is  greater  than  17576;  therefore,  7  is  too  great  for 
the^init  figure  of  the  root.  We  next  try  6,  and  find 
(26)3—17576;  therefore,  26  is  the  true  root. 

Let  the  learner  extract  the  third  roots  of  the  following 
numbers,  by  a  process  similar  to  the  preceding. 
\1.   5832./^  3.    13824. -^Z^f^  5.   50653.      ' 

^.    3375.  /<r  4.    32768. -^Jt^       6.   91125.- 

Art.  97.  Let  us  now  extract  the  third  root  of  3 
number  consisting  of  more  than  six  figures ;  for  example, 
that  of  14706125. 

Operation, 
14  706125(245.     Root. 

8 

1st  Dividend  =    67 (12 =  1st  Divisor. 

(24)3  —  13824  .  . .    ~~ 
2d  Dividend  z=     8821  . .  (1728  ,  .  =  2d  Divisor 
(245)3=14706125 

"o.  ~" 


^31.]        THIRD  ROOTS  OF  NUMBERS.  171 

Remark.     The  points  are  used  in  the  preceding  opera- 
tion, to  show  the  rank  of  the  figures  placed  before  them.    ^ 
Zeros  properly  occupy  the  place  of  these  points.     But  no 
marks    need   be    used,  provided    the   figures    are   placed 
according  to  their  proper   rank. 

Having  separated  the  number  into  periods,  we  see  that 
the  root  must  consist  of  three  figures,  viz.,  hundreds,  tens, 
and  units.  Let  a  represent  the  hundreds,  h  the  tens,  and 
c  the  units,  of  the  root. 

By  multiplication  we  find  {a -\- h -\- cY  :=.  a^ -\- ^  a^  h -\- 

3  a  62  4-  63  _^  3  ^2  c  +  6  a  6  c  +  3  62  c  +  3  a  c2  +  3  6  c2  +  c3. 

This  may  be  put  into  the  following  form, 

a3  +  3a26-|-3a62^63_^3(«_^5)2c-(-3(a  +  6)c2  + 

c^ ;  for, 

3fl2c  +  6a6c  +  362c  =  3(a2  +  2a6  +  62)c  = 

^{a-\-hfc  (Art.  59); 

and3ac2  +  36c2:=3(a  +  6)c2. 

Our  first  step  is  to  find  the  hundreds  of  the  root,  the 
third  power  of  which  is  terminated  by  six  zeros,  and  must 
therefore  be  found  in  the  14  (millions).  The  greatest 
third  power  of  hundreds  found  in  14  (millions),  is  8  (mil- 
lions), the  root  of  which  is  2  (hundreds).  We  therefore 
place  2,  corresponding  to  a,  as  the  first  figure  or  hundreds 
of  the  root,  and  subtract  8  (millions),  corresponding  to  a^, 
from  the  given  number.  The  remainder  6706125  corre- 
sponds to  3  a2  6  +  3  «  62  _^  53^  &,c. 

Our  next  step  is  to  find  6,  the  tens  of  the  root.  The 
formula  3  a2  ^  _|_  3  ^  52  _|_  53^  &bQ>.y  shows  that  by  dividing 
by  3  a^y  that  is,  by  three  times  the  second  power  of  the 
hundreds  of  the  root,  we  shall  obtain  either  the  tens  of 
the  root,  or  a  number  somewhat  too  great. 

But  three  times  the  second  power  of  hundreds  multi- 
plied by  tens,  must  be  terminated  by  five  zeros,  and  can, 
herefore,  contain   no  significant  figure  below  the   sixth 


172  EXTRACTION    OF    THE  [<§>  31. 

place  from  the  right.  Consequently,  it  is  sufficient  to 
subtract  the  8  (millions)  from  14  (millions),  the  first 
period,  and  to  bring  down,  at  the  right  of  the  remainder, 
the  sixth  figure,  that  is,  the  first  figure  of  the  next  period, 
in  order  to  form  a  dividend. 

Dividing  this  dividend  67,(6700000),  by  12,(120000), 
three  times  the  second  power  of  2  (hundreds),  we  have  5 
for  a  quotient.  But  if  we  put  5,  as  the  tens  of  the  root, 
at  the  right  of  the  2  (hundreds),  and  raise  25  (ten,*!?)  to 
the  third  power,  the  result  will  be  too  great.  We  next 
try  4,  and  find  the  third  power  of  24  (tens)  to  be  13824 
(thousands),  corresponding  to  a^  -{-  S  a^  b  -\-  3  a  b'^  -\-  b^y 
which  we  subtract  from  the  given  number,  and  have  for  a 
remainder  882125. 

This  remainder  corresponds  to  S  {a  -^  b)'^  c  -}-S  (a-^-b)  c^ 
-{-  c^ ;  and  in  order  to  find  c,  the  units  of  the  root,  we 
must  evidently  divide  by  3  {a-\-b)^,  that  is,  by  three  times 
the  second  power  of  24  (tens),  already  found,  which  is 
1728  (hundreds).  But  three  times  the  second  power  of 
tens  into  units,  must  be  terminated  by  two  zeros,  and  can, 
therefore,  contain  no  significant  figure  below  the  third 
place  from  the  right.  It  is  sufficient,  therefore,  to  sub- 
tract the  third  power  of  24  (tens)  from  the  first  two 
periods,  and  to  bring  down,  at  the  right  of  the  remainder, 
the  first  figure  of  the  next  period  to  form  a  dividend. 

This  dividend  8821  (hundreds)  divided  by  1728  (hun- 
dreds), three  times  the  second  power  of  24  (tens),  gives 
for  a  quotient  5,  which  we  place  as  the  third  or  unit  figure 
of  the  root.  Raising  245  to  the  third  power,  we  have 
14706125,  which  shows  that  245  is  the  true  root. 

Art.  98.  Hitherto,  after  having  found  a  new  figure 
of  the  root,  w^e  have  raised  the  whole  root,  so  far  as  ascer- 
tained   to  the  third  power,  and  subtracted  the  result  from 


^31.] 


THIRD    ROOTS    OF    NUMBERS. 


173 


as  many  of  the  left-hand  periods,  as  there  were  figures 
already  found  in  the  root.  But,  regard  being  paid  to  the 
local  value  of  the  figures,  the  process  of  extracting  thi' 
root  may  be  considerably  shortened.  To  show  the  ma*i' 
ner  in  which  this  is  effected,  we  shall  again  pxtract  t\  ' 
root  of  14706125. 

Operation 


14706125  ( 245. 
8 

67'06..    (12    . 
24 
16 

Root. 

.  =  3a9. 
.  —  Sab. 

1456 
4 

.iz:3a2-f3a6-f62. 
=  h. 

5824  .  .  . 

=  3a26  +  3a63_|_63. 

«S21^25  ( 1728  . .  =  3  {«  -f  6)2. 

360.  =  3(a4-^)c- 
25  =  c2. 


176425  =  3  (a  +  6)2+3  (a +  6)  c  +  c^. 
5  —  c. 


882125 
0. 


=  3(a  +  6)2c-f3{a  +  6)c2+c3 


The  process  is  the  same  as  in  the  preceding  Article^ 
until  we  have  found  the  second  figure  of  the  root ;  except 
that  we  bring  down  the  whole  of  the  second  period,  at  the 
right  of  the  first  remainder,  separating  the  two  right-hand 
figures  by  an  accent.  Previously  to  finding  6,  the  tens  of 
the  root,  we  subtracted  a^  from  the  first  period,  and  our 
remainder  with  the  succeeding  period  annexed,  6706, 
contains  6  (3  a^  -|-  3  a  6  -j"  6^),  together  with  the  thousands 
arising  from  the  rest  of  the  formula.  Having  found  6  = 
4  (tens),  we  are  to  ascertain  the  value  of  b{Sa^-\-^ab 


174  EXTRACTION    OF    THE  [^31 

-f-  6^),  and  subtract  it  from  the  dividend,  including  the  two 
figures  cut  off. 

Our  divisor  is  S  a^  =  12,  {120000),  and  3a6z=3.2.4, 
or  rather  3  .  200  .  40  =  24000,  is  three  times  the  product 
of  the  figure  last  found  by  the  preceding  figure  of  the 
root.  But  since  b  is  of  the  order  of  units  next  below  «, 
the  number  corresponding  to  Sab  v^ill  contain  a  signifi- 
cant figure  one  degree  lower  than  is  found  in  3  a^. 
Therefore,  24,  z=  3  a  6,  is  to  be  put  under  12,  =  3  a^,  one 
place  farther  to  the  right.  We  next  find  6^^  z=  (4)^==  16, 
or  rather  (40)2zz:1600,  and  since  it  contains  a  significant 
figure  one  degree  lower  than  3  a  6,  we  put  16  under  24,  = 
Sab,  one  place  farther  to  the  right  than  this  last. 

Adding  these  three  numbers,  as  the  figures  now  stand, 
we  have  1456  (hundreds),  :=  3  a^  +  3  a  6  -f"  ^^  which  we 
multiply  by  4  (tens),  =  b,  and  obtain  5824  (thousands),  := 
Sa^b-{-^ab^+b^.  We  subtract  this  product,  5824,  from 
the  dividend,  including  the  two  figures  rejected  in  the 
division,  bring  down  the  next  period  to  the  right  of  the 
remainder,  and  have  882125  =  3  (a  +  6)2  c  +  3  (a  +  b)  c^ 
+  c3i=[3(a  +  6)24-3(«  +  6)c  +  c2]c. 

Rejecting  the  two  right-hand  figures  of  8'82125,  we  take 
the  rest  as  a  dividend,  and  divide  by  1728  (hundreds),  = 
3  («  4"  ^)^y  that  is,  by  three  times  the  second  power  of  the 
hundreds  and  tens  of  the  root.  The  division  gives  5,  =  c, 
which  we  place  as  the  units  of  the  root.  We  now  wish 
to  find  the  value  of  [3  («  +  6)2  +  3  («  +  6)  c  +  c^]  c, 
and  subtract  it  from  882125. 

Our  divisor  is  3  (a  +  6)2,  nz  1728  (hundreds),  and 
3  (a  +  6)  r  =  3  .  24  .  5  =  360,  or  rather  3  .  240  .  5  =: 
3600,  is  the  product  of  three  times  the  figure  last  found 
by  the  preceding  figures  of  the  root ;  and  as  this  product 
would,  if  the  last  figure  of  it  did  not  happen  to  be  zero, 
contain  a  significant  figure  one  degree  below  the  value  of 


^81.]        THIRD  ROOTS  OF  NUMBERS.  175 

3  (a  -|-  6)2,  we  put  360  under  1728,  one  place  farther  to 
the  right  Wo  now  put  25,  =  c^,  which  contains  a  sig- 
nificant figure  still  one  degree  lower,  under  360,  one  place 
still  farther  to  the  right. 

Adding  these  numbers,  as  the  figures  now  stand,  we 
have  176425  :==  3  (a  +  6)2  +  3  («  +  6)  c  +  c^,  which  we 
multiply  by  5,  =  c,  and  have  882125  =:  3  (a  +  6)2  c  + 
S  (a -\- b)  c^  -{-  c^.  This  subtracted  from  the  last  divi- 
dend, including  the  rejected  figures,  leaves  no  remainder* 
Hence,  the  work  is  complete. 

Art.  99.  From  the  preceding  analysis  we  deduce  the 
following 

RULE    FOR    EXTRACTING    THE    THIRD    ROOTS    OF    NUMBERS. 

1.  Commencing  at  the  right ,  separate  the  number,  by 
means  of  accents y  into  periods  of  three  figures  each ;  the 
left-hand  period  may  contain  one,  two,  or  three  figures, 

2.  Find  the  greatest  third  power  in  the  left-hand 
period,  place  its  root  at  the  right,  and  subtract  the  power 
from  that  period, 

3.  To  the  right  of  the  remainder  bring  down  the  next 
period,  separating  the  two  right-hand  figures  by  an  ac- 
cent ;  those  to  the  left  of  the  accent  will  form  a  dividend. 
For  a  divisor  take  three  times  the  second  poiucr  of  that 
part  of  the  root  already  found.  Divide  the  dividend  by 
the  divisor,  and  put  the  quotient  as  the  second  figure  of 
the  root. 

4.  Take  three  times  the  product  of  the  figure  last 
found  by  the  preceding  part  of  the  root,  and  place  it 
under  the  divisor,  one  place  farther  to  the  right ;  under 
which,  one  place  farther  to  the  right,  place  the  second 
power  of  thi  figure  of  the  root  last  found.     Add  together 


176         EXTRACTION  OF  THIRD  ROOTS.       L§*3^* 

the  divisor  and  the  numbers  placed  tinder  it,  as  the  figures 
stand,  and  multiply  the  sum  hy  the  figure  of  the  root  last 
found.  Subtract  this  product  from  the  dividend,  include 
ing  the  two  rejected  figures, 

5.  To  the  right  of  the  remainder  bring  down  the  next 
period,  forming  a  new  dividend,  in  the  same  manner  as 
the  first  was  formed.  Take  for  a  divisor  three  times  the 
s^ond  power  of  the  whole  root  so  far  as  found ;  divide 
and  place  the  quotient  as  the  next  figure  of  the  root. 

6.  Find  three  times  the  product  of  the  last  figure  by 
the  whole  of  the  preceding  part  of  the  root,  and  put  it 
under  the  divisor,  one  place  farther  to  the  right ;  under 
this,  one  place  farther  to  the  right,  put  the  second  power 
of  the  last  figure  of  the  root  found.  Add  the  divisor  and 
the  numbers  placed  under  it,  as  the  figures  stand,  multiply 
the  sum  by  the  last  figure  of  the  root  found,  and  subtract 
the  product  from  the  dividend,  including  the  rejected 
figures. 

7.  Repeat  the  operations  stated  in  the  5th  and  6th  parts 
of  the  rule,  until  the  given  number  is  exhausted. 

Remark  1st.  Whenever  the  divisor  is  not  contained  in 
the  dividend,  or  the  figures  to  the  left  of  the  two  rejected, 
put  a  zero  in  the  root,  and  bring  down  the  next  period, 
separating  the  two  right-hand  figures ;  the  divisor  for 
finding  the  next  figure  of  the  root  will  then  be  the  same 
as  before,  except  with  the  annexation  of  two  zeros. 

Remark  2d.  Whenever  the  number  to  be  subtracted 
exceeds  that  from  which  it  is  to  be  taken,  diminish  the 
last  figure  found  in  the  root,  until  a  number  is  obtained 
which  can  be  subtracted. 

I.   What  is  the  third  root  of  525557943  ? 


<5>32.]  THIRD    ROOTS    OF    FRACTIONS.  177 

^^      *  525557'943(807.     Root       ""^ Hf T 

I)    ^-Oi/i       5i2_      —     (zj  /l^/&i},^^% 

1680 
49 


1936849 

7 


13557943 

0. 

Extract  the  third  roots  of  the  following  numbers 


HiXtract  tne  tnird  roots  oi  tne  lollowmg  numbers        /  n^ 
^2.    1815848.-  ^^^  7.   66430125.   ~       r^ 

C      /  3.   3652264.  ^    P"^  8.    147197952. ,y>/^^.    , 

}         4.   21024576.-4^^  ^9.    167284151^^    "5  3/. 

Cvy^S.   35937.     r     /^  ""     10.   491169069.        /^f 

/         6.    18609625.     >^^^         H.    1967221277.       j  Q  fC  ' 


SECTION  XXXII. 

THIRD    ROOTS    OF    FRACTIONS AND    THE    EXTRACTION    OF 

THIRD    ROOTS    BY    APPROXIMATION. 

Art.  100.  Since  fractions  are  multiplied  together 
by  taking  the  product  of  their  numerators  for  a  new  nu- 
merator, and  that  of  their  denominators  for  a  new  denomi- 
nator, the  third  power  of  a  fraction  is  found  by  raising 
both   numerator    ^nd    denominator   to   the   third   power. 

Thus.   (i)^  =  ^,and    (^Y  =  ^. 

'     \  7  /  343'  \  6  /  63 

12 


l73  THIRD    ROOTS    OF    FRACTIONS,    AND  [<§>  32. 

Conversely,  ihe  third  root  of  a  fraction  is  found  by 
extracting  the  third  root  of  both  numerator  and  denomi- 
nator.    For  example,  the  chird  root  of  2t  is  §,  that  of 

Find  the  third  rootg^of  the  following  fractions. 

3.   tWsV-  J^^  6.    1t¥A-  //  ^ 

Art.  101.  Bur iT  either  numerator  or  denominator  is 
not  an  exact  third  power,  we  can  obtain  only  an  approxi- 
mate root.  We  can,  however,  always  render  the  denomi- 
nator a  perfect  third  power,  without  changing  the  value  of 
the  fraction.  This  is  done  by  multiplying  both  numerator 
and  denominator  by  the  second  power  of  the  denominator. 

mu  3         3.52        3.25         75         ,  ^    .    ^  , 

1  hus,  —  =: = z=  —  :    the  nearest  mte^ral  root 

'     5         5.52  53  125'  & 

of  the  numerator  of  which  is  4,  and  the  root  of  the 
denominator  is  5.  Therefore,  f  +  is  the  approximate 
root  of  f . 

A  nearer  approximation  may  be  made,  if,  after  multi- 
plying both  numerator  and  denominator  of  the  fraction  by 
the  second  power  of  the  denominator,  we  multiply  both 
numerator  and  denominator  of  the  result  by  any  third 

75 

power.     Thus,  after  converting  f  into  — ,  we  may  multi- 

53 

75 

ply  both  numerator  and  denominator  of  —  by  8^ ;  this 

38400       ,  .       ^  .      r      1--   I,    •       34 

gives  ,  the  approximate  root  of  which  is = 

Art.  103.  The  root  of  a  whole  number,  which  is 
not  a  third  power,  may  be  approximated  in  a  similar  way; 


<5>32.]  THIRD    ROOTS    BY    APPROXIMATION.  179 

by  first  converting  it  into  a  fraction  whose  denominator  is 

1  .   T  T^  1     o       3 .  53      375    , 

an  exact  third  power.     For  example,  3  =i  — -  =:  — ,  the 

approximate  root  of  which  is  ^  -}"• 

But  the  best  mode  of  approximating  the  third  root  either 
of  a  whole  number  or  of  a  fraction,  is,  to  convert  it  into 
a  fraction,  whose  denominator  is  the  third  power  of  10, 
100,  or  1000,  &.C.  ;  that  is,  convert  it  into  lOOOths, 
lOOOOOOths,  lOOOOOOOOOths,  6lc.,  and  find  the  nearest 
root  of  the  result.  The  root  will  then  be  found  in  deci- 
mals. 

^  ,_        5.103        5000      ,  .      r      u-    u  •       1^ 

For  example,  5  =: i=  — ,  the  root  of  which  is  — 

^  103  1000  10 

-f-  =:  1*7  -\-,  If  a  more  accurate  root  is  wanted,  we  may 
change  5  to  lOOOOOOths;  thus,  5  =  |^ggg§^,  the  root 
of  which  is  j-U  —  ==  I'Tl  — , 

Hence,  it  is  evident  that  the  denominator  may  be 
omitted,  and  that  it  is  sufficient  to  annex  three  zeros  to 
the  number  for  every  additional  figure  in  the  root.  Nor 
is  it  necessary  to  write  all  the  zeros  at  once,  but  we  may 
annex  three  to  the  remainder,  when  an  additional  figure 
of  the  root  is  required,  in  the  same  manner  as  we  bring 
down  successive  periods. 

In  like  manner,  to  find  the  third  root  of  a  vulgar  frac- 
tion, we  change  it  to  a  decimal  with  thrice  as  many  deci- 
mal figures  as  we  want  decimals  in  the  root. 

When  the  number  whose  root  is  sought  contains 
whole  numbers  and  decimals,  and  the  number  of  decimal 
figures  is  not  a  multiple  of  three,  make  it  so  by  annexing 
zeros ;  or,  commencing  at  the  decimal  point,  separate  the 
whole  numbers  into  periods  by  proceeding  towards  the 
left,  and  the  decimals  by  proceeding  towards  the  right, 
and  then  complete  the  right-hand  period,  if  necessary,  by 
annexing  zeros. 


180  THIRD    ROOTS    OF    FRACTIONS.  [<§>  32. 

These  preparations  being  made,  the  root  of  a  number 
containing  decimals,  is  found  in  the  same  way  as  that  of 
an  integral  number,  care  being  taken  to  point  off  one 
third  as  many  decimals  in  the  root  as  there  are  in  the 
power,  including  the  zeros  annexed. 

1  Extract  the  third  root  of  2,  accurate  to  three  deci- 
mal figures. 

Operation, 

2-    ( 1-259  +. 
1 

lO'OO  ( 3 
6 
_4 

364 
2 

728 

272000  ( 432 
180 
25 

45025 
5 

225125 

468750^00  ( 46875 
3375 

81 

4721331 
9 

42491979 
4383021. 
Extract  the  third  roots  of  the  following  numbers,  accu- 
rate to  two  decimals. 


<§»33.1         EQUATIONS  OF  THE    THIRD  DEGREE.              181 

2.    4.  -  '<  ^^^^  6.  3-7.  10.  if. 

3     7.  7.  655.  11.  2f 

4.  9.  8.  7-75.  12.  5f. 

5.  15.  9.  f.  13.  9^. 


SECTION   XXXIIl. 

QUESTIONS  PRODUCING  PURE  EQUATIONS  OF  THE  THIRD 
DEGREE. 

Art.  103*  An  equation  of  the  third  degree  is  such 
as,  when  reduced  to  its  simplest  form,  contains  at  least 
one  term  in  which  there  are  three,  but  no  term  in  which 
there  are  more  than  three,  unknown  factors. 

When  an  equation  with  one  unknown  quantity  contains 
the  third  power  only  of  that  quantity,  it  is  called  a  pure 
equation  of  the  third  degree.  Thus,  x^  =  729  is  an  equa- 
tion of  this  kind. 

1.  In  a  package  of  cloth  there  are  as  many  pieces  as 
there  are  yards  in  each  piece,  and  it  is  worth  ^  as  maiiy 
cents  per  yard  as  there  are  yards  in  a  piece.  Required 
the  number  of  pieces  and  the  price  per  yard,  the  whole 
being  worth  $90. 

Let  X  =z  the  number  of  pieces,  also  the  number  of  yards 

in  a  piece ;  then  —  z=  the  price  per  yard  in  cents.    Hence, 


z^  =  the  whole  number  of  yards  ;  and 
.-^  =  ^  =  tV 

3  3 

We  have,  therefore, 

^  =  9000,  . 
3 

2:3  =  27000, 


;2 .  —  =  ~  =  the  price  of  the  whole  in  cents 

3  3 


182  EQUATIONS    OF    THE    THIRD    DEGREE.  [<§>  33 

x=zSO  pieces,  and  there  were  30  yards  in  a  piece. 
— .  zz:  10  cents,  price  per  yard.  ^ 

2.  The  length  of  a  rectangular  box  is  twice  the  breadth, 
and  the  de^th  is  f  of  the  breadth.  The  box  holds  200 
cubic  feet.     Required  the  three  dimensions.      ;;  ^ 

Remark.  The  cubical  contents  of  any  rectangular 
space,  or  rectangular  solid,  are  found  by  taking  the  product 
of  the  length,  breadth,  and  depth. 

3.  A  pile  of  wood  is  27  feet  long,  25  feet  wide,  and  5 
feet  high.  If  the  same  quantity  of  wood  were  in  a 
cubical  form,  what  would  be  the  length  of  one  side  of 
the  pile? 

4.  Two  numbers  are  to  each  other  as  4  to  5,  and  the 
sum  of  their  third  powers  is  5103.  Required  the  num- 
bers. /  i  "  " 

5.  What  two  numbers  are  such,  that  the  second  power 
of  the  greater  multiplied  by  the  less  makes  75,  and  the 
second  power  of  the  less  multiplied  by  the  greater  makes 
45? 

Let  X  =z  the  greater,  and  y  =  the  less.     Then, 

(1)  x2y  =  75; 

(2)  XI 
From  the  1st 


j2yzz:75;  > 

r.  1/^  =  45.     i 


_75     ^      Q__5625 

Substituting  this  value  of  y^  in  the  2d, 

5625  X         . ^ 

=  45,  or 

X* 

z=z  45.     Hence, 

45x3  —  5625;  x^=:125,.'. 

x=i5,  the  greater,  and 

V  =  —  =  —  =  3,  the  less. 


^34.]  POWERS    OF    MONOMIALS.  183 

6.  The  product  of  two  numbers  is  28,  and  8  times  the 
second  power  of  the  greater,  divided  by  the  less,  gives  98 
for  a  quotient.     What  are  these  numbers  ?     /    ^   '^ 

7.  The  sum  of  the  third  powers  of  two  numbers  is 
2728,  and  the  difference  of  those  powers  is  728.  Re- 
quired the  numbers.        I  ^  ^    /-?-  » 

8.  The  breadth  of  a  piece  of  land  is  ^  of  its  length, 
and  it  is  worth  ^  as  many  dollars  per  square  rod,  as  there 
are  rods  in  the  breadth.  The  whole  piece  being  worth 
$9375,  what  are  the  dimensions?       ^ /  5  ^^i-  ^ 

9.  A  gallon  being  231  cubic  inches,  what  is  the  length 

of  one  side  of  a  cubical  box  holding  5  gallons  ?      y  ^)^     h  -j^ 

10.  A  bushel  being  2150f  cubic  inches,  required  tHe 
side  of  a  cubical  vessel  containing  7  bushels.    J^J/,  6^  7 


SECTION  XXXIV. 


POWERS    OF    MONOMIALS. 


Art.  104:«  Any  power  of  a  quantity  is  found  by 
multiplying  that  quantity  by  itself  as  many  times,  less  one, 
as  there  are  units  in  the  exponent  of  the  power.  The 
second  power  of  a  or  a^  is  a .  a  =  a^  +  ^  =  a^,  (Art.  30)  ; 
this  is  the  same  as  a^  >^  ^, 

The  third  power  of  a^  is  a^.a^.a^=  a2+2+2  =z  a^ ; 
this  is  the  same  as  a^xa^ 

The  second  power  of  a^  53  jg  a^b^ ,  a^h^z=a^'^^b^'^^ 
=  a^b^;  this  is  the  same  as  a^  x  2  53  x  2. 

The  third  power  of  3  m^  n^  is  27  m^  n^ ;  this  is  the  same 

as  27^2X3^3X3. 

Thus  we  perceive  that,  in  all  these  examples,  we  have 


184  POWERS    OF    MONOMIALS.  [^34. 

multiplied  the  exponent  of  each  letter  by  the  exponent  of 
the  power  to  which  the  quantity  was  to  be  raised,  and  in 
the  last  example  we  actually  raised  the  numerical  coeffi 
cient  to  that  power  by  multiplication.     Hence  we  hav 
the  following 

RULE    FOR    RAISING    A    MONOMIAL    TO    ANY    POWER. 

Raise  the  numerical  coefficient  to  the  required  power, 
and  multiply  the  exponent  of  each  letter  by  the  number 
which  marks  the  degree  of  that  power. 

It  is  manifest,  moreover,  that 

Any  power  of  a  product  is  the  product  of  that  power 
of  each  of  its  factors.  For  example,  the  third  power  of 
3  a  6  c  is  27  a^  b^  c^,  which  is  the  product  of  the  third 
powers  of  3,  a,  b,  and  c. 

From  the  rule  for  the  signs  in  multiplication,  it  follows, 
that  when  the  index  of  the  power  to  which  a  quantity  is 
to  be  raised  is  an  even  number,  the  power  will  always 
have  the  sign  -f- ;  but  when  the  index  is  an  odd  number, 
the  power  will  have  the  same  sign  as  the  root.  Thus  the 
second,  fourth,  sixth,  &/C.  powers  of  any  quantity,  whether 
positive  or  negative,  will  have  the  sign  -|- ;  but  the  third, 
fifth,  &c.  powers  of  a  negative  quantity  will  have  the 
sign  — ,  while  the  same  powers  of  a  positive  quantity  have 
'  the  sign  -j-.  For  example,  the  second  power  of  -f-  «  is 
-f-  a^,  and  the  second  power  of  —  a  is  also  -f-  a^ ;  but  the 
third  power  of  a  is  -|-  a^,  while  the  third  power  of  —  a  is 

1.  Find  the  2d  power  of  3  xy^.       ^^Jf^^ 

2.  Find  the  3d  power  of  2  a^  b^.        ^  ^W  ^    V 

3.  Find  the  2d  power  of  7  a2xy2.     uxf  Od  ^    ) 

4.  Find  the  5th  power  of  2  a  62  c\     ^^  (jut^^C 

5.  Find  the  10th  power  of  az^y^.  ,/    m 


^  )^^  'O    f-Lo    -S-y 

§35.]  POWERS    OF    PotiTNOMIALS.  185^ 

6.  Find  the  mth  power  of  a^x.  Ans.  a2"*x'][  ^  , 

*     7.  Find  the  2d  power  of  —Ix^y^,     Ij  9^'l^C  '    ^^  '^  ^ 

8.  Find  the  3d  power  of —3  a3y4.   ^  lyjff^  JT 

^    9.  Findthe5thpower  of  —  2a2m3.       /^  )L       9^H 

10.  Find  the  3d  power  of  ^.       .    ^  ^  4^  f)^  ' 


10.  Find  the  3d  power  of  .  /    v 

11.  Find  the  5th  power  of 


12.    Find  the  4th  power  of  - 


~s     ^    .^  13.    Find  the  6th  power  of .  ~v_  V  .  ^    £a/  Ci 

j5^nx^     14.    Find  the  3d  power  of  a"*.  Ans.  a^r^v  ' — ^^j7^ 

/hu'tA^^  15.    Find  the  y^ith  power  of  c^h^,  c  /(h^  ^ 


SECTION   XXXV. 


POWERS    OF    POLYNOMIALS. 


Art.  103.  Any  power  of  a  polynomial  is  indicated 
by  enclosing  it  in  a  parenthesis,  or  putting  it  under  a  vin- 
culum, and,  in  both  cases,  putting  the  index  of  the  power 

3 

at  the  right.     For  example,  (a^  -|_  2  m)^,  or  a^-\-2m  , 

represents  the  third  power  of  a^-f"^^- 

Operations  may  be  performed  upon  powers  of  polyno- 
mials thus  represented,  in  the  same  manner  as  upon  the 
powers  of  simple  quantities.  Thus,  to  raise  {x--\-2i/)^  to 
the  fourth  power,  we  multiply  the  exponent  by  4,  and 
obtain  {x-\-2  y)^.  Also,  when  several  quantities  are  rep- 
resented as  multiplied  together,  the  whole  is  raised  to  any 
power,  by  raising  each  factor,  whether  monomial  or  poly- 


186  POWERS    OF    POLYNOMIALS.  ["§>  35. 

nornial,  to  the  power  required.  For  example,  the  third 
power  of  {m  —  2n){c^-^Sd)^  is  (m  —  2nf{c^  +  Sd)^; 
and  the  fourth  power  of  3  a:  {m^  —  n^)^  (x  -j-  y)^  is 
81x4(w^2  — ;i2)i2(3:-|-y)i6. 

Note.  The  learner  must  be  careful  to  distinguish  fac- 
tors from  terms.  In  the  quantity  Sm{x  —  y)^  {a^  -f-  6^)3, 
the  different  factors  are  3,  m,  x  —  y,  and  a^  -f"  ^^• 

Indicate  the  specified  powers  of  the  following  quan- 
tities, ^i- 

1.  The2dpower  of  2:  +  2y.        C^^^J^ 

2.  The  3d  power  of  {Sm-^n^f,   "  \ 

3.  The  5th  power  of  (x  +  2  6)3.      ^ 

4.  The  4th  power  of  a  (6  +  ^Y-        ^       ^  h 

5.  The2dpower  of  5x2(wi  — w)7/      ^''Y 

6.  The  2d  power  of  (a  +  6)'".   '    ^T^ 

7.  The  mth  power  of  (x  —  y)3.  ""  9^  "*  '^^  - 
^.  8.  The  6th  power  of  {a  —  hf(m^-\-n\'^  .  «f  -^  ^" 
'    9.    The  2d  power  of  3  a^  (^  —  2  y )3  (m  -f-  nf,   ^  ^ 

10.  The  3d  power,  of  ^^.     Ans.  (^^  V,  or  i^±^. 

m  —  n  \m  — n/  (m — n)' 

11.  The  4th  power  of  Jl±f.  V 

12.  The  3d  power  of  <"=->-. '  y7+^  , . 

13.  The  2d  power  of  ''"+J^<"--^'.        ^f^^A^ 


# 


Art  lOO.  If  the  powers  of  polynomials  are  required 
in  a  developed  form,  they  may  be  found  by  multiplication. 
Thus,  (3x  +  y)3=(3a:  +  3/)(3x  +  y)(3x  +  y)z=27x3 
+  27a:2y  +  92:y2_^y3. 

But  the  development  of  powers  of  polynomials  by  mul- 
tipli  nation,  when  the  powers  are  of  a  high  degree,  becomes 
very  tedious.     A  mode  has  however  been  discovered,  by 


§.35.]  POWERS    OF    POLYNOMIALS.  187 

which  any  power  of  a  binomial  may  be  developed  with 
great  facility.  The  principle  of  this  method  is  called  the 
Binomial  Theorem,  and  was  first  discovered  by  Sir  Isaac 
Newton.  It  is  particularly  adapted  to  binomials,  but  may 
be  extended  to  quantities  consisting  of  more  than  two 
terms. 

Let  us  form  some  of  the  powers  of  the  binomial  x  +  a 
by  actual  multiplication, 
(x  +  o)^  =  X  +  a, 
%  +a 


x^^ 

xa 

+ 

xa-\'a^ 

(^ 

■^ 

of. 

=  *2  +  2 

X  -\-a 

2:3  +  2 

+ 

xa-^a^. 

x2a+    xa^ 
x^a-\-2xa^ 

+  «3 

(^ 

^  + 

of 

=  s3  +  3 
X  -\-a 
z4  +  3 

x2«  +  3xa2 
x3  a  +  3  x2  a 

i  +  a3. 

xa3 

^  + 

ay. 

+ 

x3a  +  3x2a 

2  +  3 

xa3  +  ^ 

(2 

zra4  +  4 

x3  a  +  6  x2  « 

9  +  4 

xa3  +  a4.' 

X  -\-a 

z5  +  4 

x4a+    6x3 

a2  + 

4x2^3  + 

xa^ 

+ 

x^  a  +   4  x3 

«9  + 

6x2^3  +  4 

xa4 

+  a5 

(x  +  a)5^a:5  +  5x4a+10x3a2+i0a:2a3  +  5a;«4_[_^5. 

The  developments  of  the  powers  of  x  —  a  will  be  the 
same  as  those  of  x  +  a,  except  that  the  2d,  4th,  6th,  &c 
terms  will  have  the  sign  — ;  that  is,  all  the  terms  in 
which  an  odd  power  of  the  negative  term  enters,  have  the 
sign  — ,  all  the  others  having  the  sign  +.  This  neces- 
sarily follows  from  the  rule  for  the  signs  in  multiplication ; 
for,  when  the  number  of  negative  factors  is  even,  the 
product  has  the  sign  +,  but  when  the  number  of  negative 
factors  is  odd,  the  product  has  the  sign  — . 


i- 


188  POWERS    OF    POLYNOMIALS.  [<§>  35. 

We  shall,  therefore,  obtain  by  multiplication  the  first 
five  powers  of  x  —  a,  as  follows,  viz. 
{%  —  a)i  =  2;  —  «. 
(a;_a)2zzr  2:2  —  22:  «  +  «2. 
(a;_a)3,zz2:3  — 3  2:2a  +  3  2:«2_«3. 

\x  —  aY  =  x^  —  ^.%^a-\-^x^a^  —  ^xa^-^a!^, 

(2:  — «)5  =  2:5  — 5  2:4  a  +  10  2;3  a2_io  x2  a3  _|_  5  a;  «4_«5 


Art.  107.     From  an  examination  of  the  preced: 
developments,  we  shall  be  able  to  deduce  the  law,  with 
regard  to  the  letters,  exponents,  and  coefficients. 

1.  We  petceive  that  2:,  the  first  or  leading  term  of  the 
binomial,  is  found  in  every  term  of  the  development  ex- 
cept the  last ;  and  that  «,  the  second  term  of  the  binomial, 
is  found  in  every  term  of  the  development  except  the  first. 

2.  The  exponent  of  2:,  or  of  the  leading  term,  is,  in  the 
first  term  of  the  development,  equal  to  the  index  of  the 
power  to  which  the  binomial  is  raised,  and  goes  on  de- 
creasing by  unity  in  the  succeeding  terms. 

The  exponent  of  «  is  1  in  the  second  term  of  the 
development,  and  goes  on  increasing  by  unity  in  the  suc- 
ceeding terms,  until  in  the  last  term  it  has  the  same  expo- 
nent as  X  in  the  first.  Thus  the  terms  of  the  5th  power 
of  2:  -(-  «,  without  their  coefficients,  are  2:^,  x^  a,  x?  a^,  2:^  c^^ 

3.  The  coefficient  of  the  first,  as  well  as  that  of  the 
last  term  of  the  development,  is  always  1. 

The  coefficient  of  the  second  term  is  always  the  same 
as  the  index  of  the  power  to  which  the  binomial  is  raised. 
Thus  the  coefficient  of  the  second  term  of  the  develop- 
ment of  (2:  -\-  a)3  is  3 ;  that  of  the  second  term  of  the 
development  of  {x-\-ci)^  is  4,  &c. 

The  coefficient  of  the  third  term  is  found  by  multiply- 
ing the  coefficient  of  the  second  term  by  the  exponent  of 


^  35.]  POWERS    OF    POLYNOMIALS.  189 

X  in  the  same  term,  and  dividing  the  product  by  2.  The 
coefficient  of  the  fourth  term  is  found  by  multiplying  that 
of  the  third  term  by  the  exponent  of  x  in  the  same  term, 
and  dividing  the  product  by  3.  For  example,  the  second 
term  of  the  development  of  {x  -[-  «)^  being  4  x^  a,   the 

4.3 

coefficient  of  the  third  term  is  -^-1=6;    annexing  the 

letters  with  their  proper  exponents,  we  have  6  x^  a^  for  the 

third  term.     In  like  manner,  6  x^  a^  being  the  third  term, 

6 .2 
the  coefficient  of  the  fourth  term   is  -^  =  4 :    and  we 

3  ' 

have  for  the  fourth  term  4x^a. 

Thus,  if  we  multiply  the  coefficient  of  any  term  by  the 
exponent  of  x  in  the  same  term,  and  divide  the  product 
by  the  number  marking  the  place  of  that  term  from  the 
first  inclusive,  the  result  will  be  the  coefficient  of  the 
succeeding  term. 


Art.  108.  Hence,  having  one  term  of  any  power  of 
a  binomial,  the  succeeding  term  may  be  found  by  the  fol- 
lowing 

RULE. 

Multiply  the  given  term  by  the  exponent  of  x  in  that 
term,  that  is,  by  the  exponent  of  the  leading  quantity  of 
the  binomial,  and  divide  the  product  by  the  number  which 
'marks  the  place  of  the  given  term  from  the  first  inclusive; 
diminish  the  exponent  of  x  by  1,  and  increase  that  of  a 
by  1. 

This  rule,  which  admits  of  a  rigorous  demonstration, 
enables  us  to  develop  any  power  of  a  binomial. 

Let  it  be  required  to  develop  {x  -\-  a)^. 

We  know,  from  what  has  been  said,  that  the  first  term 
is  x^,  and  that  the  second  is  9  x^  a.    The  succeeding  terms 


rj^^f^nj   f^^AJ  r  ^'^J     >         ^   ' 


m  t^- 


190 


POWERS    OF    POLYNOMIALS. 


[<^35. 


may  be  found  by  the  rule,  and  the  process  of  finding  them 
IS  exhibited  below. 

The  1st  term  is  x^, 


2d 
3d 

4th 

5th 

6th 

7th 

8th 

9th 

10th 


''  9x^a, 


^xTa^=S6xra^, 


7.36 

3 

6.84 

4 

5.126 

6 
4.126 
6 
3.84 
7 
2.36 

1.9 


x^  a^  =1 84  x^  a'^, 
x^  a^  =  126  x^  cfi, 
2:4  ^5  -- 126  x4  a5, 
x^  a^  =z  84  x3  a^, 
x^a^  =  S6x^a', 
xa^  =  9xa^, 


-  a9  =:  a9. 


Hence, 
{x  +  af=ix9-}-9x^a  +  S6xTa^  +  8ix^a^'\-126x^a^  + 
UexU^-i'Six^a^  +  Sex^aT  +  dxa^'^a^, 
Remembering  that  odd  powers  of  negative  quantities 
have  the  sign  — ,  we  shall  find,  by  means  of  the  rule,  that 
the  seventh  power  of  x  —  a  is  as  follows. 
{x  —  ay  =  x^  —  7 x^ a -{-21  x^  a^—S5x^ a^  +  S5x^ a"^    ' 
—  21x^a^-{-7xa^  —  d^, 

Art.  109.  The  labor  of  developing  any  power  ot  a 
binomial  may  be  facilitated  by  attention  to  the  following 
principles. 

From  the  preceding  examples  and  the  table  of  powers 
given  in  Article  106,  we  see, 

1.    That  the  number  of  terms  in  the  development  of 


J  s) 


T     )l^ 


^ 


<5>35.]       v\  POWEKB    OF    POLYNOMIALS.  191 

any  power  of  a  binomial,  exceeds  by  1  the  index  of  that 
power.  Thus  the  development  of  {x  -{-  «)^  has  7  terms ; 
that  of  {x  —  ay  has  10  terms. 

2.  If  the  number  of  terms  in  the  development  is  odd, 
there  is  one  coefficient,  in  the  middle  of  the  series,  greater 
than  any  of  the  others ;  but  if  the  number  of  terms  is 
even,  there  are  two  coefficients,  in  the  middle,  of  equal 
value  and  greater  than  any  of  the  others.  Moreover, 
those  coefficients  which  precede  and  those  which  follow 
the  greatest  or  greatest  two,  are  the  same,  but  are  ar- 
ranged in  an  inverse  order. 

Hence,  after  half  or  one  more  than  half  of  the  suc- 
cessive coefficients  have  been  found,  the  rest  may  be 
written  down  without  the  trouble  of  calculation. 

Develop  the  following  quantities. 

1.    (a +  6)6.  4.    (m  +  /i)". 

3.    (m  —  ny.  6.    {x  +  2  7/)^ 

In  the  6th  example,  we  must  raise  the  numerical  coeffi- 
cient of  y  to  the  requisite  powers^  We  first  write  the 
development,  merely  indicating  the  powers  of  2  y.  Thus, 
(a:  +  2y)5i=zx5-f  5 2:4.2^  +  10x3  (2y)2  + 10x2  (2 y)3  + 
5x(2y)4+(2y)^ 

Now,  raising  2  y  to  the  powers  indicated,  and  putting 
the  results  instead  of  2y,  (2y)2,  &/C.,  we  have 

xS  + 10  x4y  +  40  x3y2  _[.  80  x2y3  _j^  80  a;y4  _[.  32y5. 

7.    (5x  +  «)4.  8.    (a2-f.x2)5. 

To  develop  the  8th,  we  first  indicate  the  powers  of  a* 
and  x2.     Thus, 

(a2  -j-  2:2)5  —  (a2)5  ^  5  («2)4  2-2  _[_  10  (a2)3  (2,2)2  _f. 
10  (a2)2  (z2)3  _j.  5  (^2)  (a:2)4  _^  (^2)5. 

Note      In  this  example,  a^  is  the  leading  term,  and  the 


192  ROOTS    OF    MONOMIALS.  [^36. 

exponents  of  a^  outside  of  the  parentheses  are  to  be  used 
in  finding  the  coefficients. 

Now,  raising  a^  and  x^  to  the  powers  indicated,  and 
substituting  the  results,  we  have 
(a2 -j_  2:2)5  —  «io  _|_  5  ^8  a;2  _|_  10  a6  a;4  ^  10  a4  2-6  +  5  «2  a;8 

+  xio. 

As  an  example  in  which  the  binomial  theorem  may*  be 
used  to  fir^'  any  power  of  a  quantity  consisting  of  more 
than  two  te'  ms,  we  shall  develop  (a  -|-  6  +  c)^. 

Substitute  any  single  letter,  x,  for  instance,  instead  of 
b  -\-  c]  then  a  -f-  6  -f-  c  becomes  a-\-x. 

Now,  (a  +  a:)4  — a4-f4a3a;-(-6a5a;2_|_4«a;3_|_a;4^ 

But  x^=:h-\-c,  .'. 

xS  zn  ( 6  +  c )  2  izz  62  _j_  2  6  c  +  c2  ; 

a:3z=(6  +  c)3  =  63  +  362c  +  36c2+c3;  and 

x4i=(&  +  c)4=z:64_|^4  63c-f6  62c24-4  6c3-}-c4. 

Putting  these  values  instead  of  x,  x^,  &/C.,  we  have 
a4+4a3(6  +  c)  +  6a2(62_|_2  6c  +  c2)-f4«(63-|-3^62c 
+  3  6c2  +  c3)  +  ^^  +  4  63c  +  6  62c24-4  6c3  +  c4. 

Lastly,  performing  the  multiplication  indicated,  we  have 
(a  +  6  -f  c)^  =  «^  +  *  «^  ^  +  4  «3  c  +  6  «2  62  _j.  12  ^2  5  c 

+  6a2c2  +  4«63_|_i2«62c  +  12a6c2  +  4«c3  +  54 

+  463c  +  662c2  4-4  6c3  +  c4. 


SECTION  XXXVI. 


ROOTS     OF     MONOMIALS. 


Art.  110.  From  the  manner  in  which  a  monomial  is 
raised  to  iy  power,  shown  in  Article  104,  results  the  fol- 
lowing 


§36.]  ROOTS    OF    MONOMIALS  193 

RULE    FOR    EXTRACTING    THE    ROOT    OF    ANY    MONOMIAL. 

Extract  the  root  of  the  numerical  coefficient y  and  divide 
the  exponent  of  each  literal  factor  hy  the  number  which 
marks  the  degree  of  the  root. 

The  reason  of  this  rule  is  manifest,  as  may  be  shown 
by  an  example.  Thus,  the  second  power  of  3  a:  y^  is 
9  3;i  X  2  y2  X  2  ^:3  9  2:2  y4  .   consequently,  the  second  root  of 

2      4 

9  x^  1/"^  =:S  x^  7/^  =:  S  X  y^.  In  like  manner,  the  third  root 
of  27x6  y9  is  3x^y^i=:3x2y3. 

It  is  to  be  remarked. 

That  every  root  of  an  even  degree  may  have  either  the 
sign  -f-  or  — ;  but  a  root  of  an  odd  degree  has  the  same 
sign  as  the  power. 

Thus,  the  4th  root  of  -f"  ^^  na^y  be  either  -|-  a  or  —  « ; 
because  (+  a)^  =  +  a^,  and  ( —  a)^  =  -|-  «^.  Whereas 
the  third  root  of  -f-  a^  is  --|-  a,  but  the  third  root  of  —  a^ 
is  ^-  a ;  because  (-f-  «)^  =  +  ^^>  but  ( —  a)^  =  —  a^. 

We  have  already  said,  (Art.  86,)  that  the  second  root 
of  a  negative  quantity  is  imaginary.  The  same  may  be 
said  of  any  even  root  of  a  negative  quantity.  Thus,  the 
fourth  root  of  — 81,  and  the  sixth  root  of  — a  are  imagu 
nary  quantities, 

1.  Find  the  second  root  of  a^h^, 

2.  Find  the  second  root  of  Ax^y^. 
5*    Find  the  second  root  of  9  a^  x^  y^, 

^.    Find  the  third  root  of  a^  W  c^. 

5.  Find  the  third  root  of  27  2:3  3^9  2;i2 

6.  Find  the  third  root  of  —  125  a^  x^, 

7.  Find  the  fourth  root  of  d^  W>  x^'^, 

8.  Find  the  fifth  root  of  x5yio. 

0    Find  the  second  root  of . 

16  1/4 

iO.    Find  the  third  root  of  ^-i^^. 
13 


194  ROOTS    OF    MONOMIALS.  [<§>  36. 

Art.  111.  From  the  preceding  examples,  also  from 
what  was  shown  in  Article  104,  relative  to  any  power  of 
a  product,  we  infer  that 

Any  root  of  a  product  is  the  product  of  the  roots,  to 
the  same  degree,  of  each  of  the  factors  of  that  product. 

For  example,  the  third  root  of  27  a^  6^  is  3  a  6^,  which 
is  the  product  of  the  third  roots  of  27,  a^,  and  h^,  the 
factors  of  27  a^  h^. 

In  a  similar  manner,  if  any  numerical  quantity  is  sepa- 
rated into  factors  that  are  exact  powers  of  the  required 
degree,  which  may  always  be  done  when  the  number 
itself  is  an  exact  power  of  that  degree,  we  may  extract 
separately  the  root  of  each  factor,  and  afterwards  multiply 
these  roots  together. ,  Thus,  1296  =  9  .  144,  the  second 
root  of  which  is  3  .  12  izi  36. 

Art.  113.  Since,  in  extracting  the  root  of  a  mono- 
mial, we  divide  the  exponent  of  each  letter  by  the  number 
expressing  the  degree  of  the  root,  it  follows,  that  if  any 
exponent  is  not  divisible  by  that  number,  the  division  must 
be  expressed,  and  this  gives  rise  to  fractional  exponents 

1  2 

For  example,  the  third  root  of  a  is  a^,  that  of  a^  is  a^. 

The  expression  a^  represents  either  the  third  root  of  a^, 

1  1  ^  V  2         ^ 

or  the  second  power  of  a^;   for   {a^Y  z=z  a^  "^     =  a^. 

3. 

Also,  a^  denotes  either  the  fifth  root  of  a^,  or  the  third 

power  of  a^. 

The  radical  sign,  as  well  as  fractional  exponents,  may 
be  used  to  indicate  a  root  of  any  degree,  provided  we 
place  over  this  sign  a  number  expressing  the  degree  of  the 

root.     Thus  Y     ,  which  is  the^same  as  i/    ,  indicates 

3    —  .  4  — 

the  second  root  *,    i/    ,  the  third  root ;   i/    ,  the  fourth 


<§>37.]  SECOND    ROOTS    OF    POLYNOMIALS.  195 

root.     Hence,  we  have  the  following  equivalent  expres- 
sions, viz. 

1         3 2        4  3         6  X 

i/  m  =  w^" ;  U  m^zzLit^ \  ^ m^=im^ \  ^ nO :=z m^ ,  &c. 
We  may,  therefore,  use  indifferently  either  the  radical 
sign  or  a  fractional  exponent,  remembering  that  the  num- 
ber over  the  radical  sign  is  the  denominator,  and  that 
the  exponent  of  the  quantity  under  the  sign  is  the  nu- 
merator, of  the  fractional  exponent. 


SECTION  XXXVII. 

SECOND  ROOTS  OF  POLYNOMIALS. 

Art.  113.     It  is  required  to  extract  the  second  root 

of  25x2  +  60  2:^  +  36  2/2. 

Operation, 
25  x2  +  60  a;  y  +  36  y2  (  5  a;  +  6  y .     Root. 
25x2 

60 xy  +  36y2  (I0a;  +  6y 

60xy  +  36y2 

o; 

By  comparing  this  quantity  with  a^-^-^  ah  -\-lfi,  the 
second  power  of  a-\-h,  and  recollecting  the  process  of 
extracting  the  second  roots  of  numbers,  we  shall  readily 
see  the  mode  of  proceeding.  The  first  term  25  x^  corre- 
sponds to  a2  J  Yv^e  therefore  extract  the  root  of  25  x^, 
which  is  52:,  (Art.  110,)  place  it  at  the  right,  and  sub- 
tract its  second  power  from  the  given  quantity.  / 

The  remainder  60  a:  y  +  36  y2,  which  we  regard  as  a 
dividend,  corresponds  to  2a6  +  62,  or  {2a-\-b)h,     Di- 


196  SECOND    ROOTS    OF    POLYNOMIALS.  [<§>  37 

viding  the  first    term  60  x  y  corresponding  to  2  a  6,  by 
10  X  corresponding  to  2  a,  we  have  6  y  answering  to  b. 

We  now  place  6y,  with  its  proper  sign,  in  the  root 
also  at  the  right  of  our  divisor,  and  have  l0x-\-6i^ 
answering  to  2  a  -f-  6.  Multiplying  10  2;  -|-  6  y  by  6  y,  we 
obtain  60  a:  y  -(-  36  y^  corresponding  to  {2a-\-  b)  b.  Sub- 
tracting this  product  from  the  dividend,  we  have  no  re- 
mainder.    Consequently,  5a:-f-6y  is  the  required  root. 

When  a  quantity  consists  of  more  than  three  dissimilar 
terms,  its  second  root  will  consist  of  more  than  two  terms. 
But  the  process  of  finding  the  second  root  of  a  polynomial 
is,  in  all  cases,  so  similar  to  that  of  extracting  the  root  of 
a  number,  that  it  hardly  needs  a  separate  explanation 
The  following  example  will  serve  as  an  illustration. 

What  is  the  second  root  of  9  a^  —  24  a^  6  _[_  22  a^  b^  — 
8a63  +  6M 

Operation. 
9  ^4  _  24  a3  5  ^  22  a2  62  _  8  a  63  -f  64  {Sa^  —  ^ab  +  b^. 

9a^ 

—  24  a3  6  +  22  a2  62  —  8  a  63  ^  64  {6a^  —  4ab 
--24a36-fl6a262 

6a262_8a63-f-64(6a2_8a6  +  6Q 
6a2  62  — 8a63^64 

0. 

The  process  of  finding  the  first  two  terms  of  the  root  is 
precisely  the  same  as  in  the  first  example  of  this  article. 

Having  obtained  the  second  dividend,  6a^b'^  —  Sab^ 
-j-  6^^  we  double  the  first  two  terms  of  the  root,  and  have 
for  a  second  divisor  6a^  —  8  a  6. 

Performing  the  division,  we  obtain  6^  for  the  third  term 
of  the  root,  which  we  annex,  with  its  proper  sign,  both 
to  the  preceding  part  of  the  root  and  to  the  divisor.  Our 
divisor  then  >ecomes  6  a^  —  Sab  -\-b'^,  which  we  multiply 


^37.]      SECOND  ROOTS  OF  POLYNOMIA.S.        197 

by  6^,  and  subtract  the  product  from  the  second  dividend. 
As  there  is  no  remainder,  the  root  required  is  Sa^  —  4  «  6 
-|-  62,  or  4  «  6  —  3  a^  —  6^,  for  the  second  power  of  eithei 
will  produce  the  given  quantity. 

Art.  114:.  From  the  foregoing  examples  and  expla- 
nations, we  derive  the  following 

RULE     FOR     EXTRACTING     THE     SECOND     ROOT     OF     A 
POLYNOMIAL. 

1.  Arrange  the  quantity  according  to  the  powers  of 
some  letter, 

2.  Find  the  root  of  the  first  term,  and  place  it  as  the 
first  term  of  the  root  sought ;  subtract  the  second  power 
of  this  term  from  the  given  polynomial,  and  call  the  re- 
mainder the  first  dividend. 

3.  Double  the  term  of  the  root  already  found  for  a 
divisor,  by  which  divide  the  first  term  of  the  dividend, 
and  place  the  quotient,  with  its  proper  sign,  as  the  second 
term  of  the  root,  also  at  the  right  of  the  divisor.  Multi- 
ply the  divisor,  with  the  term  annexed,  by  the  second  term 
of  the  root,  and  subtract  the  product  from  the  dividend. 

4.  The  remainder  will  form  a  second  dividend,  which 
is  to  be  divided  by  twice  the  whole  root  found,  and  the 
quotient  is  to  be  placed,  as  the  next  term  of  the  root,  also 
at  the  right  of  the  divisor.  Multiply  the  divisor,  with 
the  term  last  annexed,  by  the  last  term  of  the  root,  and 
subtract  the  product  from  the  last  dividend. 

5.  The  remainder  will  form  a  new  dividend,  with  which 
proceed  as  before ;  and  thus  continue,  until  all  the  terms 
of  the  root  are  found. 

Remark  1.  As  we  at  first  arrange  the  given  polyno- 
mial according  to  the  powers  of  some  letter,  so  the  same 
arrangement  must  be  preserved  in  each  dividend. 


198       SECOND  ROOTS  OF  POLYNOMIALS.     [*§>  37 

Remark  2.  In  dividing,  we  merely  divide  the  first  term 
of  the  dividend  by  the  first  term  of  the  divisor ;  and  it  is 
manifest,  from  the  manner  in  which  the  divisors  are  ob- 
tained, as  well  as  from  inspection,  that  the  successive 
divisors  will  have  their  first  terms  alike. 

Extract  the  second  roots  of  the  following  quantities. 

1.  9m2  +  24;?^c+16c2. 

2.  25x2  +  70xy  +  49y2. 

3.  3662_486a:  +  162:2. 

4.  x4— 4a;?/3  +  3/4_42;3y_|_62;2^2. 

5.  9a:4  +  30a:3y  +  25a:2y2_42a:2_702;i/  +  49. 

6.  12x5  +  .5a:44-4a;6  +  72;2  — 22;3  — 2a;  +  l. 

7.  25  2:4  3/4_70  2;3y3-f  49x2y2. 

8.  9  m6  —  12  7/25  +  34  m4  _  20  7;i3  _|_  25  m^., . 

Art.  lis.  The  following  additional  remarks  may  be 
found  useful, 

1.  No  hinomial  can  he  an  exact  second  'power ;  for  the 
second  power  of  a  monomial  is  a  monomial,  and  the 
second  power  of  a  binomial  necessarily  contains  three 
terms.  Thus,  x^  -|-  y^  cannot  be  an  exact  second 
power.  It  wants  -^-^xy  to  make  it  the  second  power 
of  a:  -f-  3/,  and  it  wants  —  2  x  y  to  make  it  the  second 
power  of  X  —  y, 

2.  In  order  that  a  trinomial  may  be  a  perfect  second 
power,  it  must  be  such,  that,  when  it  is  arranged  accord- 
ing to  the  powers  of  a  particular  letter,  the  extreme  terms 
shall  both  be  positive,  and  shall  both  be  exact  second  pow- 
ers, and  the  mean  term  shall  be  twice  the  product  of  the 
second  roots  of  those  powers. 

When  these  conditions  are  fulfilled,  the  second  root 
may  be  found  in  the  following  manner. 

Extract  the  second  roots  of  the  extreme  terms,  writing 
these  ^oots  after  each  other,  and  giving  them  both  the  sign 


^38.]  IRRATIONAL    QUANTITIES.  199 

-[-,  when  the  second  term  of  the  trinomial  is  positive,  but 
giving  one  of  them  the  sign  — ,  when  that  second  term  is 
negative.  The  result  will  be  the  second  root  of  the  tri- 
nomial. 

For  example,  49  x^ — 112  a:  y  -f-  ^  y^  is  an  exact  second 
power.  The  root  of  the  first  term  is  7  x,  that  of  the  third 
term  is  8y,  and  twice  the  product  of  these  roots  is  112xy. 
But  since  1 12  a:  y  in  the  given  trinomial  has  the  sign  — , 
the  required  root  is  7  x  —  By,  or  8  ?/  —  7  x. 


SECTION   XXXVIII. 

TRANSFORMATION    AND    SIMPLIFICATION    OF    IRRATIONAL 
QUANTITIES. 

Art.  116.  We  have  already  seen,  in  Article  86,  that, 
when  a  quantity  is  not  an  exact  power  of  the  same  degree 
as  the  root  required,  this  root  is  expressed,  either  by  the 
radical  sign  or  by  fractional  exponents.  Such  expressions 
are   in  general  called  irrational   quantities.     Thus,  l/2, 

i/  a,  Y  (1%  w  a  b^,  or  the  equivalent  expressions  2^,  a^, 

a^,  d^b^,  are  irrational. 

We  have  also  seen,  (Art.  110,)  that  to  extract  the  root 
of  a  monomial,  we  divide  the  exponent  of  each  factor  by 
the  number  expressing  the  degree  of  the  root ;  so  that, 
when  this  division  can  only  be  represented,  it  gives  rise 
to  fractional  exponents.  But  any  root  of  a  quantity  is 
also  expressed  by  writing  the  quantity  under  the  radical 
«ign,  and  putting  the  number  denoting  the  degree  of  the 


200     TRANSFORMATION    AND    SIMPLIFICATION    OF    [^38. 


3 2.      1     4  4  

root  over  the  sign.    Thus,  i/a^ bc'^^za^  b^ c^,  and  i/a^xy^ 

=:  a^  x^  y^.  Hence,  we  may  readily  convert  a  quantity 
having  a  radical  sign  into  an  equivalent  expression  with 
fractional  exponents,  in  the  following  manner. 

Remove  the  radical  sign,  and  divide  the  exponents  of  the 
different  factors  under  the  sign,  hy  the  number  placed  or 
supposed  to  be  placed  over  it ;  for  that  number  denotes 
the  degree  of  the  root  required. 

Transform  the  following  expressions  into  equivalent 
ones  having  fractional  exponents. 

1.   i/a6.        Ans.  (ab)^  =:a^b^ 
3.   ^^. 


4.   ^a^b^c. 


5.  y/m  x^  y. 

6.  ^/^362. 

7.  ^^+6.       Ans.  {a-\-b)i. 

8.  ^a^ix  +  y),     Ans.  a*(x  +  y)* 

9.  \/ia-i-b)xy. 


10.    wm  x^  y^ 


11.    ^m^{x  —  y). 


12.    y^m^y^a  +  b). 

Art.  117,  On  the  other  hand,  it  is  manifest  that 
expressions  with  fractional  exponents  may  be  converted 
into  equivalent  ones  with  the  radical  sign,  by  a  reverse 
operation,  as  follows. 

Take  away  the  denominators  to  the  fractional  exponents 


^38.]  IRRATIONAL    QUANTITIES.  201 

of  the  different  factors,  supposing  them  to  have  a  common 
denominator,  placing  the  result  under  the  radical  sign,  and 

3      7  4 

putting  the  denominator  over  it.     Thus,  dS  h^  =z  yd^  W, 

If  the  fractional  exponents  have  not  all  the  same  de- 
nominator, they  must  be  reduced  to  a  common  denomina- 
tor ;  and  integral  exponents,  if  there  are  any,  must  be 
converted  into  fractions  with  the  same  denominator  ;  after 
vi^hich  proceed  as  before. 

Thus,    «*  h^  =  a^  h^^  =  '^osTio  ;    and   x^  y^  z^  = 

12     2      3  6 

X^  y^  Z^  =:  i/x^2  y2  2;3. 

Remark,  Taking  away  the  denominators  of  the  frac- 
tional exponents  of  all  the  factors,  after  all  the  exponents 
have  been  reduced  to  a  common  denominator,  is  equiva- 
lent to  raising  the  quantity  to  the  power  denoted  by  that 
denominator. 

Transform  the  following  quantities  into  equivalent  ex- 
pressions with  the  radical  sign. 


1. 

x*y* 

6. 

(^h^C. 

2. 

7. 

1    \ 

3. 

f   i 

8. 

xy^zi. 

4. 

x^y^. 

9. 

«^(^+y)  • 

5. 

a*6*c2. 

10. 

{a-\-b)i{x-y)i. 

Art.  118.  We  have  shown,  in  Article  111,  that  the 
root  of  a  product  is  formed  by  multiplying  together  the 
roots  of  all  the  factors  of  that  product.  Hence,  we  may 
extract  the  roots  of  all  such  factors  as  are  exact  powers 
of  the  requisite  degree,  and  indicate  the  roots  of  the  other 
factors. 

Let  it  be  required  to  find  the  second  rcKDt  of  32  a^  h^. 


202     TRANSFORMATION    AND    SIMPLIFICATION    OF    [<§>  38. 

This  root  is  indicated  thus,  \/Wc^,  or  thus,  (32a3  65)i 
But  32«365z=  16. 2a264^j  — 1(5^2^4. 2a6.  Now,  16, 
a^,  and  lA  are  exact  second  powers.  We  may,  therefore, 
take  the  roots  of  these  factors,  and  place  their  product  as 
a  coefficient  to  the  expression  for  the  second  root  of  2  a  6. 
We  then  have  ^32 a^ h^  =  ^/Wa^b'^.^ab  =i4ab^ ^2ab 

=  4ab^{2ab)i. 

In  a  similar  manner  we  have  (81  a^  b^)^  =l  (27  a^  b^)^  X 

(3  a  b^)i  =  3ab{a  b^)i  =Sab  ^'ab^.  In  this  case,  we 
find  all  the  factors  which  are  exact  third  powers. 

In  order  to  separate  an  irrational  quantity  into  factors, 
for  the  purpose  of  simplifying,  we  seek  the  greatest 
numerical  factor  that  is  an  exact  power  of  the  degree 
required,  and  the  greatest  exponent  of  each  literal  factor, 
not  exceeding  its  given  exponent,  that  is  divisible  by  the 
number  which  denotes  the  degree  of  the  root. 

For  example,  in  the  expression  ( 128  a''' 6^)^,  although  4 
and  16  are  factors  of  128,  and  are  exact  second  powers, 
yet  the  greatest  factor  of  128,  which  is  also  an  exact 
second  power,  is  64.  The  exponent  of  a  being  7,  the 
greatest  number  less  than  7,  and  divisible  by  2,  is  6 ;  also 
the  greatest  number  divisible  by  2,  and  not  exceeding  6,  the 
exponent  of  b,  is  6  itself.  Hence  128  d^  b^,  when  resolved 
into  the  requisite  factors,  becomes  64  a^b^  .2  a,  the  second 

root  of  which  is  8  a^  b^  i/2  a,  or  8  a^  b^  (2  a)^. 
Simplify  the  following  expressions. 

1.  (a^bf.  5.   y/'lOS. 

2.  \/^T^  6.    (56a3a:5)2. 


3.  ^8a^x\  7.    (72  a5) 

4.  (27a4  2:3)*  8.    2^; 


^38.]  IRRATIONAL    QUANTITIES.  203 


3.- 


9.    a^Scfib^c.  12.   ^^3  (2  a  +  6). 

10.  3^/686«3  6.  .        13.    ^2a:2  +  3x2y. 

11.  (3a:3m4)*  14.    (3  «3  _|.  4  ^4)* 

Art.  119.  When  the  quantity  which  is  under  the 
radical  sign,  or  which  is  enclosed  in  a  parenthesis  with  a 
fractional  exponent,  is  a  fraction,  the  expression  may  be 
simplified  in  the  following  manner,  viz. 

Multiply/  the  numerator  and  denominator  of  the  Jrac- 
tion  hy  such  a  quantity  as  will  render  the  denominator 
an  exact  power  of  the  requisite  degree ;  then  take  the  roots 
of  the  denominator  and  of  such  factors  of  the  numerator 
as  are  exact  powers. 

—  j2^,  we  multiply  both  numerator 
and  denominator  of  the  fraction  by  3  6;  the  expression 
then   becomes    (   ^   ^  W  z=  (— .  3g6l^.      Taking   the 

second  root  of  the  fraction  —  aiTd  placing  it  as  a  factor 
before  the  parenthesis,  we  have  —  (3  «  6)^. 

3  6 

In  like  manner  M*=  (^\^=  (-i-  .35«2)i= 

\25  63/  \125  63/  \I25  63  / 

—  (35«2)* 
"-    Simplify  in  a  similar  manner  the  following  expressions. 

MB)*-  "•»©*• 


204  IRRATIONAL    QUANTITIES  [^38. 

Art.  ISO.  As  we  can  extract  the  root  of  any  factor, 
and  place  it  as  a  factor  before  the  radical  sign  or  the 
parenthesis  with  a  fractional  exponent,  so  we  may  put 
under  the  sign,  or  within  the  parenthesis,  any  factor 
standing  before  it,  if  we  first  raise  that  factor  to  a  power 
of  the  same  degree  as  the  root. 

Thus,  Sai/  b=^^9  a^b;  to  obtain  this,  we  raise  3  a  to 
the  second  power,  and  place  the  result  as  a  factor  under 

the  radical  sign.    In  like  manner,  %m{x y)^  i=  f -^— ^ J ^ ; 
to  obtain  this,  we  raise  —  to  the  third  power,  and  multiply 

o 

X  y  within  the  parenthesis  by  the  result. 

We  may,  in  a  similar  manner,  reduce  any  rational 
quantity  to  the  form  of   an  irrational   quantity.     Thus, 


2=y/4   or   4^1=^8    or    8^,  &c. ;    ah  =  ^a^h^   or 
(a2  52)i-  — ^^3j3  or  (a3 63)'S",  &c. 

Reduce  the  following  quantities  entirely  to  an  irrational 
form. 

1.  ah\/'x,  5.  3y/^m2. 

2.  2w»i/6c.  6.  4m(xy)^. 

3.  l^-a.  7.  ^(«2)* 

4.  I^v/^.  8.  i(a  +  6)t 

9.    Reduce  2  to  the  form  of  a  second  root. 

Ans.  i/4,  or  4^. 

10.  Reduce  a  6  to  the  form  of  a  second  root. 

11.  Reduce  2  a  to  the  form  of  a  third  root. 

12    Reduce to  the  form  of  a  third  root. 

c 


^39.]  IRRATIONAL    Q,UANTITIES.  205 


SECTION  XXXIX. 

OPERATIONS  ON  IRRATIONAL  QUANTITIES  WITH  FRACTIONAL 
EXPONENTS. 

Art.  131.  duantities  having  fractional  exponents 
ate,  in  general,  to  be  treated  in  the  same  manner  as  if 
vhe  exponents  were  whole  numbers. 

1.  Add  Sai  and  7  a*. 

The  sum  is  3  a*  +  7  a^  =  10  a*.    (Art.  33.) 

2.  Addax^  and3  6a:i 

The  sum  is  «  x^  +  3  5  X*  z=  (a  4-  3  b)  x*.     (Art.  59.) 

3.  From  10  a^  subtract  3  a^. 

The  difference  is  10  a^  _^  3  a*  zz:  7  a*. 

4.  From  5ax^  i/^  subtract  ^mx^  y^. 

X    2  12  4.    ^ 

The  result  is  ^ax^  y^ — 37?ix^y^=(5a — 3m)a;^y^. 

5.  Add  3  (12)*  and  4  (27)*. 

The  sum  expressed  is  3  (12)*  +  4  (27)*. 
But  by  simplifying  these  terms,  the  result  may  be  ob- 
tained in  a  reduced  form.     For,  3  (12)*  =  3.4*.  3*  = 
3.2.3*  =  6.3*,  (Art.  118;)    and  4  (27)*  =  4  .  9*  .  3* 
zz:  4  .  3  .  3*  zn  12  .  3*.      Therefore,  3  (12)*  +  4  (27)*  = 

6  .  3*  + 12  .  3*  z=  18  .  3*,  the  result  in  its  most  simple 
form 

6.  From  (250  2;3)*  subtract  (54  y3)*. 

The  difference  expressed  is  (250  x^Y  —  (^^  V^^ 


206  IRRATIONAL    Q,UANTITIES  [<§>  39. 

But  (250a:3)*i=(125a:3.2)*=:5x.2*;  and  (54y3)J 

=  (27  y3  .  2)*  1=  3  y  .  2*       Hence,  (250  a;3)*  —  (54  y3)i 

—  5a:.2*  — 3y.2*=:(5a;  — 33/)2^,   the   result   in  its 
simplest  form. 

From  the  preceding  examples  we  derive  the  following 

RULE     FOR     ADDING     AND     SUBTRACTING     IRRATIONAL 
QUANTITIES. 

Expires s  the  addition  or  subtraction  as  usual  by  signs, 
simplify  the  terms  if  possible,  and  reduce  similar  terms. 

Remark,  Irrational  quantities,  expressed  by  means  of 
fractional  exponents,  are  called  similar,  when  the  factors 
having  fractional  exponents  are  alike  in  all,  and  have  re- 
spectively the  same  exponents.  Thus,  S  ax^  and  m  x^ 
are  similar;  but  Sd^x  and  mx^  are  not  similar. 

Art.  133.     1.    Multiply  a^  by  a^. 

This  is  done  by  adding  the  exponents.     Thus,  a^ .  a* 
zzz«^+?=z=at     (Art.  30.) 

2.  Multiply  2  m'^  x^  by  3  n^  x^. 

The  product  is  6  Ttfi  '  *x^  '    °'  z=:  6  n^  x^. 

3.  Multiply  5  a^  by  4  a*. 

In  this  example,  in  order  to  add  the  exponents,  we 
must  reduce  them  to  a  common  denominator.     We  then 

have  5  «^  =  5  d^ ,  and  4  a^  =  4  a^ ;    hence,  5  «^ .  4  a^  = 

5«^.4«'^zz:20a*. 

4.  Multiply  2  (^  b^  by  7  a*  6*. 

In  this  case,  the  exponents  of  a  and  those  of  b  must 


<§>39.]  WITH     FRACTIONAL    EXPONENTS.  207 

be  separately  reduced  to  a  common  denominator.  Then 
2  (^  h^  becomes  2  a^  b^^,  and  1  a^ h^  becomes  7  a^  h^^. 
Hence,  2  «*  6*  .  7  a*  &*  =  2  a^  6**  .  7  a^  h^^  =  U  J  b^i. 

5.  Divide  m^  by  /»*. 

This  is  done  by  subtracting  the  exponent  of  the  di- 

mi         X 5. 

visor  from  that  of  the  dividend.     Thus,  —  =  w^      ^  = 

6.  Divide  6m^xi  by  2m^x*. 

In  this  case  we  reduce  the  exponents  of  the  same 
letter  in  each  quantity  to  a  common  denominator.     We 

.       ,          6m* ic*       6m^rA  i     i       ,  .  ,    .q  . 

then  have  — -— ^  = — -  =  3  m^  x^^,    (Art.  48.) 

2  m^  x^       2  m^  x^^ 

7.  Divide  3  6*  c  by  4  6^c2. 

.       .     36*c       Sb'^^c      36^"^    ,,       ^^, 
The  quotient  is  — -—  = —  = .  (Art.  63  ) 

4  6^  c2      4  b^^  c^        4  c 

2 

8.  Required  the  second  power  of  2  m^. 

This  is  performed  by  raising  the  coefficient  to  the 
second  power  and  multiplying  the  exponent  of  m  by  2. 

(Art.  104.)     Thus,  (2  m*)2  =  4  m* 

9.  Required  the  third  power  of  3  a^  b^, 

Ans.  27  «^  6^. 
Conversely,  to  find  the  root  of  an  irrational  quantity, 
we  either  extract  or  express  the  root  of  the  numerical 
coefficient,  and  divide  the  exponents  of  the  other  factors 
by  the  number  whicn  marks  the  degree  of  the  root. 

XO.   What  is  tie  second  root  of  4 a^ ?         Ans.  2 a*. 


208  IRRATIONAL    QUANTITIES  [<§>  39. 

11.   What  is  the  third  root  of  5  aH^  1 

Ans.  5*«*6i 

From  what  precedes  we  infer  that  the  following  opera- 
tions, viz.,  multiplication,  division,  Jinding  powers,  and 
extracting  roots,  are  performed  upon  quantities  with  frac- 
tional  exponents,  in  the  same  manner  as  if  the  exponents 
were  integral.  This  is  manifest ;  for  there  is  no  reason 
why  exponents  in  a  fractional  form  should  not  be  subject 
to  the  same  law  as  those  in  an  integral  form. 

Art.  1S3.  We  have  assumed,  in  what  precedes,  that 
no  change  is  made  in  the  value  of  irrational  quantities  by 
reducing  the  fractional  exponents  to  a  common  denomi- 
nator. This  is  manifestly  the  case,  since  reducing  to  a 
common  denominator  does  not  change  the  value  of  frac- 
tions, but  only  their  form. 

Hence,  we  may  reduce  the  exponents  of  all  the  factors 
m  an  irrational  quantity  to  a  common  denominator,  with- 
out changing  the  value  of  the  quantity.     Thus,  2  x^"  y^  zz: 

It  is  evident  also  that  fractional  exponents  may  be  con- 
verted into  the  decimal  form,  and  used  in  that  form  as  well 

1  3 

as    any  other.      Thus,   oF  1=  a^'^ ;    a^  =  a^"'^  ;    and   the 
X      3 

product   a^  .  «^  zz:  «0-5  ,  ^0-75  :::^  ^1 


25 


1.  Add  (8)*  and  (32)*  Ans.  6  (2)* 

2.  Add  (27)*  and  (75)*. 

'   3.  Add  (135  a)*  and  (40  a)* 

4.  Add  (250  a2)i  and  (128  a^)* 

5.  Add  (192  a3x)*  and  (24a3a;^^ 


«§>39.]  WITH    FRACTIONAL    EXPONENTS.  209 

^,    6.  From  (75)*  subtract  (48)* 

-     7.  From  3  (50)*  subtract  (18)* 

8.  From  (320«2)i  subtract  (40  aS)^ 

9.  From  (8)*  subtract  2  (J-)* 

10.  From  (54  63)*  subtract  (24  a^)*  -j;;    ^       ^ 

1 1 .  Multiply  3  X*  y^  by  5  x*  3^*. 

12.  Multiply  2  X  y*  by  3  a:*  y^, 

13.  Multiply  3  (8)*  by  2  (6)* 

The  product  is  6  .  8*  .  6*  =1  6  (8  .  6)*  =  6  (16  .  3)*  = 
6.  4  (3)*  =  24  (3)* 

14.  Multiply  4  .  5^"  by  3  .  8^,  and  simplify. 

^    "  15.  Multiply  8  (108)*  by  5  (4)*  and  simplify. 

16.  Divide  m^  by  m^, 

17.  Divide  a^  z^  by  a^  x^. 

18.  Divide  aF  by  a^. 

19.  Divide  4  x  y^  by  2  x^^^. 

20.  Divide  8  (27)*  by  4  (3)*. 

,       .    8(27)*      2(27)*      2(9.3)*      6.3* 
The  quotient  is  — ^ — j-  =  — ^^ — j-  =  — ^^ — ~-  = . 

4  (3)*  (3)*  (3)*  3* 

=  6. 

In  another  way;  --i-^  =  2  (-2^)*  ==  2 .  9*  =  2 . 3  ==  d 
4(3*) 

21.  Divide  4  (512)^  by  2  (3)*. 

22.  Find  the  2d  power  of  a*  6* 

14 


210  IRRATIONAL    Q,UANTITIES  [^4C. 

23.  Find  the  2d  power  of  2  'j^  y^ , 

24.  Find  the  3d  power  of  3  m^ x y^, 

25.  Find  the  2d  power  of  {a  —  a;)^. 

26.  Find  the  3d  power  of  5  (w  +  n)^ 

27.  Extract  the  2d  root  of  a^ 

28.  Extract  the  2d  root  of  4  a^  6^. 

29.  Extract  the  2d  root  of  9  c^  h^. 

30.  Extract  the  3d  root  of  2  a^  h^. 

Remark.     Represent  the  3d  root  of  2  in  this  example, 

31.  Extract  the  2d  root  of  3  (a+  6)^. 

32.  Extract  the  3d  root  of  40  a^  {a  —  x)^. 


SECTION   XL. 

OPERATIONS    UPON    IRRATIONAL    QUANTITIES    WITH    THE 
RADICAL    SIGN. 

Art.  1S4:«  Since  irrational  quantities  with  the 
radical  sign  may  always  be  converted  into  equivalent 
expressions  with  fractional  exponents,  (Art.  116,)  all 
operations  might  be  performed  upon  them  in  this  latter 
form. 

But  as  the  radical  sign  is  used  in  many  mathematical 
works,  we  shall  show  how  to  treat  irrational  quantities 
expressed  by  means  of  this  sign. 

Irrational  quantities  with  the  radical  sign  are  commonly 
called  radical  quantities.     The  mode  of  simplifying  irra- 


^40.]  WITH    THE    RADICAL    SIGN.  211 

tional  quantities  in  both  forms  has  already  been  shown  in 
Article  118. 

The  addition  and  subtraction  of  irrational  quantities 
with  the  radical  sign,  are  manifestly  performed  in  the 
same  manner  as  if  fractional  exponents  were  used. 
(Art.  121.) 

We  observe,  however,  that 

Irrational  quantities  with  the  radical  sign  are  said 
to  he  similar,  when  the  indices  over  the  sign  are  alike, 
and  the  quantities  under  the  sign  are  in  all  respects  the 
same. 


3 

m 


Thus,  i/«  h  and  3  i/a  h  are  similar ;  also,  y/a^  h  c  and 

3 —  3  — 

^a^  h  c  are  similar.     But  ya  h  and  ^«  h  are  not  simi- 

3 3 

lar ;  neither  are  i/a^  h  m  and  i/a^  W-  m. 
1.    Add  ^288  and  3^8. 
The  sum  expressed  is  ^288  +  3  ^8.     But  ^288  = 
12^2,    and   3^/8z=6^2.      Hence,  ^288  +  3^/8  = 
12^2  +  6^2^:18^2. 


2.    Add  m  y/27  a  x  and  3  n  ^125  a  i\ 


The  sum   is  my/27  ax -\-Sni/125ax,     But  when 

3  — 
simplified,  these  terms  become,  respectively,  Smyax  and 

3  —  ...  .  3  — 

15  n  i/«  X.    Hence,  the  sum  in  its  simplest  form  is  Smi/ax 

3 3  —  * 

-j-  15  w  t/a  a;  z=:  (3  m  +  15  n)  ya  x, 

3.    From  y/48  subtract  ^27. 

Expressing  the  subtraction,  and  simplifying,  we  have 
^48  —  ^27  =  4  ^3  —  3  ^3  =  ^3,  the  result  in  its  sim- 
plest form. 


212  IRRATIONAL    Q,UANT1TIES  [<§>  40 

3  3  - 

-       ■       -  V' 

3.- 

we    have    5  m 

3.- 


4.    From  5  m  1/8 1  x  subtract  2  n  ^24  x. 

In   this    case    we    have    5  m  1/8I  x  —  2  w  i/24  a;  zz: 
3  —  3 —  3  — 

15  m  1/3  a;  —  AnySx  =1  (15  m  —  4tn)^Sx. 


Art.  ISS^.  We  shall  deduce  rules  for  other  opera- 
tions on  radicals,  from  the  modes  given  in  the  preceding 
section,  for  corresponding  operations  on  irrational  quanti- 
ties vv^ith  fractional  exponents. 

The  following  principle  will  be  of  frequent  use,  viz. 

The  exponents  of  all  the  factors  under  the  radical  sign 
and  the  index  over  the  sign,  may  both  he  multiplied  or 
divided  hy  the  same  number  without  affecting  the  value  of 
the  expression. 

Remark.  A  numerical  factor  under  the  radical  sign 
may  either  be  considered  as  having  an  exponent,  or  it 
may  be  actually  raised  by  multiplication  to  the  power 
denoted  by  the  number  by  which  the  exponents  of  the 
literal  factors  are  multiplied,  or  the  root  may  be  extracted 
when  the  exponents  of  the  letters  are  divided. 

• —        2  — 
Thus,  in  the  expression  ^m^  or  i/m^,  we  may  multiply 

8 

the  2  and  3  both  by  4,  for  example,  which  gives  wm^'^ ;  for, 
^^  =:m^  =  m'^^'  =:  ^^.     ( Arts.  1 16,  1 17. ) 
In  like  manner  ^Wc^  =  \jWa^  =  ^Ta\ 

8 12  3  2 

On  the  other  hand,  i/m^^  ^:^  jjfw  _ -  ^2^  -—  i^jn^^  or  i/m^ 

8  

which  might  have  been  obtained  from  wm^^,  simply  by 
dividing  the  8  and  12  both  by  4. 

In  like  manner,  ^27 a^ b^  =  v/(27)* ab^=i ^/3a^,  or 
^/37P. 


<§>40.]  WITH    THE    RADICAL    SIGN.  213 

Art.  136.  Upon  the  principle  explained  in  the  fore- 
going Article,  two  or  more  radical  expressions  may  be 
made  to  have  the  same  index  over  the  sign  vs^ithout  affect- 
ing the  value  of  these  expressions. 

.  —  3  —  6  — 

Thus,  \/m  and  i/x^  are,  respectively,  the  same  as  ^m^ 

6. — 

and  \/x^.  The  first  of  these  is  obtained  by  multiplying  2 
supposed  to  be  over  the  sign,  and  the  exponent  of  m  under 
it  by  3,  and  the  second  is  obtained  by  multiplying  the  3 
and  2  both  by  2. 

In  like  manner,  i/a^ h^  and  \/a m^  are,  respectively,  the 

12 12 

same  as  i/«^  6^  and  i/a^  m^. 

This  process  is  equivalent  to  reducing  the  correspond 
ing  fractional  exponents  to  a  common  denominator,  the 
indices  over  the  sign  being  considered  as  denominators, 
and  the  exponents  under  the  sign,  as  numerators.  The 
common  index  will  therefore  be  either  the  product  of  all 
the  indices  over  the  sign,  or  their  least  common  multiple. 

Art.  137.     1.    Multiply  s^'a  by  ^m. 

The  product  is  ^a  m  ;  for  ^a  =z  a^,  and  y/m  =z  m^ ; 
hence,  i/a  .  i/m  :=a^  m^  ={a  m)^  =z  i/a m, 

2.  Multiply  7  ^a2  by  3  ^^^3. 

We  first  render  the  indices  over  the  radical  sign 

3 —  1 2 —  4  — 

alike.      We   then   have   7  ya^  =  7  ^a^,   and   3  i/m^  =z 

12 3 —  4 12 —  12 — 

3  ^m^  ;  consequently,  7  ^a^  .  3  ^rrt^  =  7  ya^  .  3  ym^  = 
7  a^^ .  3 m^s-  z=  21  d^"^  m^s-  —  21  (a^  rri^y^  =  21  \/d 

3.  Divide  9  s^a  m  by  3  i/a. 

9\/am 


8  rn^^ 


Representing  the  division,  we  have 


3\/fl 


214  IRRATIONAL    QUANTITIES  [^40. 

But,  9  i/a  m=:9  d^  m^,  and  3  i/a  =  3  a^ ;  hence, 
9[/am      9a^mi      ^    4      q    /- 
3v/a  3  a* 

4.    Divide  7  ^«  by  4  ^P. 

Rendering  the  indices  over  the  radical  sign  alike, 
and  representing  the  division,  we  have 

4/62      4v/64 

From  an  examination  of  the  results  in  the  four  pre- 
ceding questions,  vi^e  deduce  the  following 


RULE      FOR     THE     MULTIPLICATION     AND     DIVISION     OF 
RADICALS. 

Make  the  indices  over  the  radical  sign  alike,  if  they  are 
not  so ;  then  multiply  or  divide  one  coefficient  hy  the  other  ; 
also  take  the  product  or  quotient  of  the  quantities  under 
the  radical  sign ;  place  the  latter  result  under  the  common 
sign,  before  which  write  the  product  or  quotient  of  the  coef- 
ficients previously  found. 


Art.  128.     1.    Find  the  3d  power  of  3i^a^x^. 

4:   3  2  ^4:- 


Since   3^a^  x^  =  3  a^  x^,  we    have    {3^a^x^f  = 
{3a^xif  =  3^.ai^^x^^^,  (Art.   104,)  =27 a* 2;*  = 


4  - 


'9  r.6 


27  ^a^ 

4 

This  result  might  have  been  obtained  from  3  i/a^  x^,  bj 
raising  the  coefficient  3  to  the  third  power,  and  multiply- 
ing the  exponent  of  each  factor  under  the  radical  sign  by 
3,  the  number  which  marks  the  degree  of  the  power  re- 
quired. 


^40.]  WITH    THE    RADICAL    SIGN.  215 

6 . 

2.    Find  the  2d  power  of  lya^m. 

Since   7  ^a^  m  ==  7  a^  w^,   we   have    (7  ^/a^  m)^  =: 

(7  a^m^f  =  72  .  a^  ^  ^  m*  ^  ^  =  49  «*  m*  =  49  y^^s^. 

This  result  is  the  same  as  would  have  been  produced, 
if  we  had  merely  raised  the  coefficient  7,  of  the  given 
quantity,  to  the  second  power,  and  divided  the  index  over 
the  sign  by  2. 

From  these  results  we  derive  the  following 

RULE    FOR    RAISING    A    RADICAL    TO    ANY    POWER. 

Raise  the  coefficient  of  the  radical  to  the  power  required^ 
and  either  raise  the  quantity  under  the  radical  sign  to  the 
same  power,  or  divide  the  index  over  it  by  the  number  ea> 
pressing  the  degree  of  the  power. 

Art.  139.  The  process  of  extracting  a  root  is  mani- 
festly just  the  reverse  of  that  by  which  a  power  is  found ; 
hence,  we  have  the  following 

RULE  FOR  EXTRACTING  ANY  ROOT  OF  A  RADICAL. 

Extract  or  express  the  root  of  the  coefficient  of  the 
radical,  and  either  extract  the  root  of  the  quantity  under 
the  radical  sign,  or  multiply  the  index  over  it  by  the  nuni' 
ber  expressing  the  degree  of  the  root. 


For  example,  the  third  root  of  64  t/a^  x^  is  4  i/a^  x^ ; 


12- 


also,  the  fourth  root  of  2  ^m^  x  is  ^2  .  \/m^  x  =z  1/23  X 

12 12 12 

^m^x=^y2^.m^x=:^Sm^x,     This  is  the  result  in  its 

simplest  form,  although  the  answer  given  directly  by  the 

4-     12 

rule  would  be  i/2  .  i/m^  x. 

Art.    130.      Let  the  learner  perform  the  following 
questions,  simplifying  the  results  when  possible. 


216  IRRATIONAL    QUANTITIES.  ["^  ^^ 

1.  Add  ^32  and  ^18. 

2.  Add  ^25x  and  ^16^. 


3.  Add2^49ax  and  3^36aa; 

4.  Add  ^27a  and  ^Ma. 

5.  Add  3  ^250^  and  7^/^54^. 

6.  From  ^500"  subtract  ^125. 

7.  From  5  ^12  a:  subtract  ^18  x. 

8.  From  3^1  subtract  2^^. 

9.  From5y^T6  subtract  2^54. 

10.  From  2^/jm>a  subtract  ^/^24q. 

11.  Multiply  ^2  by  ^2. 

12.  Multiply  5^3  by  4y/3. 

13.  Multiply  2  ^a  by  3^/^. 

14.  Multiply  S^a  by  5^a. 

15.  Multiply  5  ^3  by  7^8. 

16.  Multiply  4  +  ^2  by  4— ^2. 

17.  Divide  i/m  x  by  i/m, 

18.  Divide  6 y^oTc  by  S^ab, 

19.  Divide  10^108  by  5^/12 

20.  Divide  ^/5  by  ^5. 

3 5  

21.  Divide  4  ^a  a;  by  2  ^a  x, 

22.  Divide  5  ^a2^  by  3y/^m. 

3 

23.  Find  the  2d  power  of  ya  m, 

24.  Find  the  3d  power  of  2 ^3«, 


'§►41.]  PROPORTIONS.  21t 

25.  Find  the  2d  power  of  ^a^  h^. 

26.  Find  the  3d  power  of  ^\/m^ 


2. 

10- 


27.  Find  the  5th  power  of  2  a:  y  ya^  b^ 

28.  Extract  the  2d  root  of  25y/a^m^. 

29.  Extract  the  3d  root  of  27  y^a6^. 

3 

30.  Extract  the  2d  root  of  16  ya  x. 

2 

31.  Extract  the  3d  root  of  ^wmy. 


32.    Extract  the  2d  root  of  9  ^4  m!^  y^. 


33.  Extract  the  3d  root  of  125  \/(a  +  hf 

34.  Extract  the  2d  root  of  144  ^«  +  6. 


SECTION  XLI. 


RATIO    AND    PROPORTION. 


Art.  131.  The  ratio  of  two  quantities  is  the  quotient 
arising  from  the  division  of  one  by  the  other,  whether  that 
division  can  be  exactly  performed,  or  whether  it  can  only 
be  expressed.  It  is  sometimes  called  ratio  by  division,  or 
geometrical  ratio,  to  distinguish  it  from  the  difference  of 
two  quantities,  which  is  called  ratio  by  subtraction,  or 
arithmetical  ratio.  But  when  the  word  ratio  simply  is 
used.  It  signifies  ratio  by  division. 

The  most  proper  way  of  expressing  a  ratio  is  in  the 
form  of  a  fraction.     Thus,  |  is  the  ratio  of  5  to  '^y  and 

—  is  the  ratio  of  m  to  w. 


i 
218  PROPORTIONS.  ['§►41. 

A  proportion  is  an  expression  of  equality  between  two 
equal  ratios.  Sometimes  the  term  geometrical  proportion 
is  used  to  express  the  same  thing.     For  example,  f  =  ^, 

and  —  zi:  —  are  proportions. 

For  the  sake  of  convenience,  two  dots,  thus  : ,  placed 
between  the  quantities,  are  used  to  express  division,  and 
four  dots,  thus  : : ,  are  used  instead  of  the  sign  i=.  Thus, 
«  :  6  :  :  c  :  c?  is  read  "  a  is  to  &  as  c  is  to  c?,"  and  has  the 

same  meaning  as  —  z=:  — .     The  signification  in  both  cases 

is,  that  a  divided  by  h  gives  the  same  quotient  as  c  divided 
by  d.  In  this  work  we  shall  sometimes  use  the  points  to 
denote  division,  but  shall  always  prefer  the  sign  =  to 
express  equality. 

In  any  proportion  a :  b  =2  c  :  d,  the  quantities  a,  6,  c, 
and  d  are  called  the  terms  of  the  proportion.  The  two 
quantities  a  and  b  are  the  terms  of  the  first  ratio ;  c  and  d 
are  the  terms  of  the  second  ratio. 

In  the  proportion  a\b:=.  a  d,  the  two  quantities  a  and 
c  are  called  the  antecedents,  and  the  two  quantities  b  and 
d  are  called  the  consequents  of  the  proportion ;  a  is  the 
antecedent  of  the  first  ratio,  and  c  that  of  the  second ;  b 
is  the  consequent  of  the  first  ratio,  and  d  that  of  the 
second.  Moreover,  a  and  d  are  called  the  extremes,  b  and 
c  the  means  of  the  proportion. 

These  names  are  expressive  of  the  position  in  which 
the  quantities  stand  with  respect  to  each  other,  when  the 
division  is  indicated  by  dots.  The  word  antecedent  signi- 
fies going  before,  and  consequent  means  following  afi;er. 
Thus,  in  the  ratio  a:b,  a  goes  before  or  stands  first,  and 
b  follows  after  it.  Also,  a  and  d  are  called  extremes,  be- 
cause they  occupy  the  ends  or  extremities  of  the  propor 


^41.]  PROPORTIONS.  219 

tion ;  b  and  c  are  called  the  means,  because  they  occupy 
the  middle  place  in  the  proportion. 

Art.  133.  We  shall  now  proceed  to  demonstrate 
those  properties  of  proportions,  which  are  most  important 
and  of  most  frequent  use. 

(i).    Take   any  proportion   a:b=zc:  d.     This    is    the 

same  as  —  =  — ,  and  if  we  multiply  by  the  denominators 
6         d 

b  and   d,  we  have  «c?=6c.     But  a  and  d  are  the  ex- 
tremes, and  b  and  c  are  the  means.     Hence, 

In  any  proportion,  the  product  of  the  means  is  equal  to 
the  product  of  the  extremes. 

(ii).    Suppose  we  have  the  equation  ad=bc.     If  we 

divide  both  members  by  b  and  d,  we  have  —  =:  — ,  or 

,  b  d 

aibzi^c  :  d.     Therefore, 

If  the  product  of  two  quantities  is  equal  to  the  product 
of  two  other  quantities,  the  tivo  factors  of  one  product 
may  be  made  the  means,  and  the  two  factors  of  the  other 
product,  the  extremes  of  a  proportion, 

(hi).  If  any  three  terms  of  a  proportion  are  known 
quantities,  we  can  always  find  the  value  of  the  remaining 
term. 

For  take   any  proportion,   a  :  b  =z  c  :  d.     This   gives, 

by   (i),   ad=zb  c  ;    hence,  by  division,   a  =z  — ,  d  =  — , 

y        ad  a  d       ^^ 

o  =  -—,  c  =  -— .     Hence, 

In  any  proportion,  either  mean  is  equal  to  the  product 
of  the  extremes,  divided  by  the  otper  mean ;  and  either  ez^ 
treme  is  equal  to  the  product  of  the  means,  divided  by  the 
ether  extreme. 


220  PROPORTIONS.  [^41 

From  this  we  infer  that, 

If  three  terms  of  one  proportion  are  respectively  equa* 
to  the  three  corresponding  terms  of  another  proportion,  tie 
remaining  term  of  one  must  he  equal  to  the  remaining  term 
of  the  other, 

(iv).  The  proportion,  a:b  =:b  :  c,  in  which  the  two 
mean  terms  are  alike,  is  called  a  continued  proportion. 
The  term  b,  in  this  case,  is  called  a  mean  proportional 
between  a  and  c,  and  c  is  called  a  third  proportional  to  a 
and  6.     From  this  proportion  we  have  b'^=i  ac,  .•.  6  = 

^ac.     Hence, 

The  mean  proportional  between  two  quantities  is  equal 
to  the  second  root  of  their  product. 

From  this  it  follows  that, 

If  the  second  power  of  any  quantity  is  equal  to  the 
product  of  two  other  quantities,  the  first  quantity  is  a 
mean  proportional  between  the  last  two. 

For,  by  (ii),  the  equation  b'^=zac  gives  a:b-=.bx  c. 

(v).    Suppose  we  have  the  proportion  a:  b=:c  :  d,    (1). 
This  gives,  by  (i),  ad=:bc. 

Now,  by  (ii),  the  equation  ad=ibc  may,  besides  the 
given  proportion,  be  converted  into  the  four  following,  viz. 
a:  c  =  b:d,{2): 
dib=c'.a,  (3)  : 
c  :  d=:a:  b,  (4) 
b  :  a=:d:  c,  (5). 
By  comparing  proportions  (2),  (3),  (4),  and  (5)  with 
the  given  proportion  (1),  we  infer  that. 

In  any  proportion,  the  means  may  exchange  places ;  the 
extremes  may  exchange  places ;  the  extremes  may  be  made 
the  means,  and  the  means  the  extremes ;  both  ratios  may^ 


^4].]  pROPOKTioNs.  ^M     Vy'      ^W^ 

at  the  same  time,  be  inverted,  that  is,  the  antecedent  ana^ 
consequent  of  each  ratio  may  exchange  places. 

(vi).  Since  a  ratio  is  a  fraction,  and  since  the  value  of 
a  fraction  is  not  changed,  when  both  numerator  and  de- 
nominator are  either  multiplied  or  divided  by  the  same 
quantity,  it  follows  that, 

In  any  proportion,  we  may  multiply  or  divide  both  terms 
of  either  ratio  by  the  same  quantity,  and  we  may  multiply 
or  divide  all  the  terms  of  a  proportion  by  the  same  quan^ 
tity,  without  disturbing  the  proportion. 

We  may  also  multiply  or  divide  both  terms  of  the  first 
ratio  by  one  quantity,  and  both  terms  of  the  second  ratio 
by  another  quantity,  or  we  may  multiply  both  terms  of  one 
ratio  by  any  quantity,  and  divide  both  terms  of  the  other 
ratio  by  the  same  or  a  different  quantity,  without  disturb^ 
ing  the  proportion, 

(vii).  Both  of  the  antecedents,  or  both  of  the  consequents, 
of  a  proportion,  may  either  be  multiplied  or  divided  by  the 
same  quantity,  without  disturbing  the  proportion. 

The  reason  is  obvious;  for,  by  multiplying  the  ante- 
cedents or  dividing  the  consequents,  we  multiply  the  ratios 
or  fractions ;  and  by  dividing  the  antecedents  or  multiply- 
ing the  consequents,  we  divide  the  ratios  or  fractions 
(Arts.  ^Q,  58.)  But  if  equal  quantities  are  both  multi- 
plied or  both  divided  by  the  same  quantity,  the  results 
must  be  equal. 

^/\ 
(viii).    Suppose  we  have  the  two  proportions,  ^  ^<:^     v 

aib=ic'.  d,  and  a:  b  =:m:  n,  Q. 

the  ratio  a  :  b  being  found  in  both  proportions.    By  Ax.  7, 

we  have 

c  :  d  =1  m  :  n.     Hence, 

Tf  two  proportions  have  a  common  ratio,  or  a  ratio  in 


222  PROPORTIONS>    ;  ^^^^^<§>41. 

one  proportion  equal  to  a  ratio  in  the  other,  the  two  re* 
maining  ratios  are  equal,  and  may  form  a  proportion. 

(ix).    Let  there  be  given  the  two  proportions  \  .fv. 

a  :  b  =z  c  :  d,  and  a  :  m  =z  c  :  n,  ,      •,? 4M^^.;. 

in  which  the  corresponding  antecedents  are  alike.  By 
changing  the  means  in  each,  according  to  (v),  the  propor- 
tions become 

,.     a  :  c=^b  :  d,  and  a:  c=zm  :  n;  hence,  by  (viii),  ,k 

b  :  d  =z m  :  n,  or  b  :  m=^ d :  n.  ./  £ 

But  b,  d,  m,  and  n  are  the  consequents  of  the  given 
proportions.     Hence,  v 

If  in  two  proportions  the  antecedents  are  alike  or  equal, 
the  consequents  will  form  a  proportion. 

Suppose  now  that  we  have  the  two  proportions 
a  :  b  =z  c  :  d,  and  m  :  b  =z  n  :  d, 
in  which  the  corresponding  consequents  are  alike. 
By  (v),  these  proportions  become 

a:  c=i  b  :  dj  and  m  :^=zb  :  d. 
Consequently,  by  (viii), 

a  :  c  =z  m  :  n,  or  a  :  m  =1  c  :  n. 
But  a,  c,  m,  and  71  are  the  antecedents  of  the  given 
proportions.     Hence, 

If  in  two  proportions  the  consequents  are  alike  or  equal 
the  antecedents  will  form  a  proportio*^ 

(x).    Suppose   we   have   the     >ioportion    a  :  b=:c  :  d 

which  is  the  same  as  —  =  '  . 
h         a 

Adding  db  1  to  eact  member,  we  have 

Reducing  each  member  wholly  to  a  fraction, 

adtzb       czbd  ,    r     r  t    j    j 
=: ,  or  adb o  :  ozzci^a  :a, 


-^^. 


%M.\       ^^^^      PROPORTIONS.       u       ,  rfisa 

which  by  (v)  becomes  n 

a±b:cThd=ib:d=a:c,  (1),  ' '-AA.Mj 

since,  from  the  given  proportion,  these  last  two  ratios  are 
equal. 

If  we  take  the  given  proportion,  invert  the  ratios,  so 

that  it  becomes  —  =  — ,  and  thi^n  proceed  as  above,  we 
shall  obtain  ' " 


% 


6zt«:^dbc=:a:c^=^5:e?.     (2).  ;,;       % 

Comparing  proportions  (1)  and  (2),  which  are  essen- 
tially alike,  with  the  given  proportion,  we  infer  that 

In  any  proportion,  the  sum  or  difference  of  the  Jirst  two 
terms  is  to  the  sum  or  difference  of  the  last  two,  as  the 
first  term  is  to  the  third,  or  as  the  second  is  to  the  fourth. 

(xi).  From  proportion  (1)  given  above,  by  taking  the 
sign  -{-,  we  have 

^■^hxc-\-d-=-hid.     By  taking  the  sign  —  in  (1), 
a  —  6:  c  —  d=:b:  d.     Hence,  by  (viii), 
a-|-6:  c-\-d=a  —  b  :  c  —  d,  or  by  (v), 
a-^b  :  a  —  b  =z  c  -{-d:  c  —  d. 
Comparing  the  last  two  proportions  with  the  original 
proportion  a :  b  =:  c  :  d,  we  infer  that. 

In  any  proportion,  the  sum  of  the  first  two  terms  is  to 
the  sum  of  the  last  two,  as  the  difference  of  the  first  two 
terms  is  to  the  difference  of  the  last  two  ;  also,  the  sum  of 
the  first  two  terms  is  to  their  difference,  as  the  sum  of  the 
last  two  terms  is  to  their  difference. 

Remark,  If  we  had  taken  proportion  (2)  in  (x),  we 
might  have  obtained  from  it 

b-^-aid-^cz^b  —  aid —  c,  and 

6-f-«:6  —  azzid-^-czd —  c, 
so  that  the  principle  stated  above  is  entirely  general. 


224  PROPORTIONS.  'v5^  [<§>41 


(xii).  Ify  in  any  proportion y  the  antecedents  are  alike 
or  equal,  the  consequents  must  be  equal;  also,  if  the  con* 
sequents  are  alike  or  equal,  the  antecedents  must  be  equaL 

The  reason  is  plain ;'  for  equal  fractions  having  equal 
numerators,  must  have  equal  denominators ;  and  equal 
fractions  having  equal  denominators,  must  have  equal 
numerators. 

Moreover,  it  is  evident  that, 

If,  in  any  proportion,  the  second  term  is  greater  than 
the  first,  the  fourth  must  be  greater  than  the  third,  and 
conversely ;  and  if  the  first  two  terms  are  equal,  the  last 
two  must  also  be  equal, 

(xiii).    Suppose  we  have  a  series  of  equal  ratios,  as 
aib:=.ci  d:=ie  :f=zg  :  h,  or 
a  c  e  g 

Let  q  represent  the  value  of  each  of  these  fractions 
Then, 

.-d         ^      a  c  e  g 

Removing  the  denominators, 

a  =  bq,  c  =  dq,  e=fq,  g  =  hq. 
Adding  these  equations, 

a^c  +  e+g={b  +  d+f+h)q. 
Dividing  by  6  +  ^  "{-/-{-  ^> 

=ig  =  —  =  — ,  &c. :  or 

a^C'-\-e"\-g:b-{-d-]-f'{'h=a:b=.c:d=^e:f=g:h. 
Now,  the  first  term  of  this  proportion  is  the  sum  of  the 
antecedents,  and  the  second  is  the  sum  of  the  consequents, 
of  the  given  ratios.     Hence, 

Ik  any  series  of  equal  ratios,  the  sum  of  the  antecedents 


^41.]  PROPORTIONS.  225 

is  to  the  sum  of  the  consequents,  as  any  one  of  the  ante* 
cedents  is  to  its  consequent. 

(xiv).   Suppose  we  have  the  two  proportions 
a  :  b  =^  c  :  d,  and 

These  are  the  same  as  «; 

±=±  '' 

f  h' 

By  multiplying  together  the  corresponding  members  of 

these  two  equations,  we  obtain 

ae        eg-    ...    . 
—  =  -^,  that  IS, 
bf      dh'  ' 

aei  hf=i  eg  :  dh. 

This  proportion  is  the  same  as  we  should  have  obtained 
from  multiplying  together  the  corresponding  terms  of  the 
two  given  proportions  in  their  first  form.  This  is  called 
multiplying  the  proportions  in  order;  and  it  is  evident 
that  any  number  of  proportions  might  be  combined  in  the 
same  way.     Hence, 

If  two  or  more  proportions  are  multiplied  in  order,  the 
result  will  form  a  proportion. 

Since  division  is  the  reverse  of  multiplication^  folic 
that, 

If  proportions  are  divided  in  order,  the  r^uit  wilt  form 
a  proportion,  ^         ^ 

(xv).    Given  a:b:=c:  d,  * " 

Putting  this  proportion  in  the  form  of  —  =:  4",  and 

b  a 

raising  both  members  to  any  power,  the  degree  of  which 
is  denoted  by  m,  we  have 

^  ==  ^ ,  or  a*^  :  6*^  zzr  c*"  :  d^.     Hence, 

IS 


226  PROPORTIONS.  ['§>41. 

Similar  powers  of  proportional  quantities  form  a  pro* 
portion. 

Since  extracting  roots  is  the  reverse  of  finding  powers, 
it  follows  that 

Similar  roots  of  proportional  quantities  will  form  a 
proportion. 

Art.  133.  The  following  exercises  are  designed  to 
exemplify  the  foregoing  principles  of  proportions.  The 
correctness  of  any  proportion  may  be  verified  by  ascer- 
taining that  the  product  of  the  means  is  equal  to  that  of 
the  extremes. 

1.  Illustrate  (i)  by  the  proportion  7  :  10  =:  21 :  30. 

2.  Illustrate  (ii)  by  putting  12.8  =  32.3  into  a  pro- 
portion ;  also  by  forming  a  proportion  from  mnz=.xy. 

3.  Illustrate  (iii)  by  finding  the  value  of  x  in  each  of 
the  following  proportions. 

x:7i=9:21; 
10:x=:5:15; 
7 :  4  =  x  :  20 ; 
3  :  5  z=  7  :  x. 

4.  According  to  (iii),  what  is  to  be  inferred  respecting 
%  and  y  in  the  proportions 

3:  7=  12:  a;  and 
3:7=12:y'? 
and  what  are  the  values  of  x  and  y  1 

5.  According  to  (iv),  what  is  the  mean  proportional 
between  5  and  20  ?  Also,  what  is  to  be  inferred  from  the 
equation 

x^  =  m{d-{-h)1 

6.  In  the  proportion 

5  :  7  =  15  :  21 

make  all  the  changes  authorized  by  (v). 


§>41.]  PROPORTIONS.  227 

7.    Illustrate  (vi)  by  the  proportion 
10  :  15  =  30  :  45. 

8  Illustrate  (vii)  by  the  proportion 

30  :  49  =  60  :  98. 

9  Illustrate  (viii)  by  the  two  proportigns 

7  :  9  ==  21 :  27,  -^ 
7:9=14:18. 

10.  Illustrate  (ix)  by  the  two  sets  of  proportions 

(  10:7  =  30:21, 
(  10:5  =  30:15; 
{  8:5=16:10, 
(  12:5  =  24:10. 

11.  Illustrate  (x)  by  the  proportion 

3:7  =  9:21. 

12.  Illustrate  (xi)  by  the  proportion 

12:8  =  60:40. 

13.  According  to  (xii),  what  is  to  be  inferred  from  the 
proportion 

9:x  =  9:3? 
also,  from  the  proportion 

y:7  =  5:7? 

14.  Also,  according  to  (xii),  what  is  to  be  inferred 
with  regard  to  x  in  each  of  {he  proportions 

4  :  10  =  12  :  X, 
6:    6  =  20:a:? 

15.  Illustrate  (xiii)  by  the  equal  ratios 

1 :  2  =  3  :  6  =  4  :  8  =  9  :  1 8  =  1 2  :  24 

16.  Illustrate  (xiv)  by  the  two  proportions 

3:    5  =  21:35, 
12:20  =  42:70. 

17.  Illustrate  (xv)  by  the  proportion 

4:9  =  36:81. 


228  PROGRESSION    BY    DIFFERENCE.  [^42 


SECTION  XLII. 

PROGRESSION    BY    DIFFERENCE. 

Art.  134:.  A  progression  by  difference^  or  an  arith^ 
metical  progression,  is  a  series  of  quantities  constantly 
increasing  or  constantly  diminishing  by  a  common  dif- 
ference ;  and  these  successive  quantities  are  called  the 
terms  of  the  progression. 

Thus,  1,  2,  3,  4,  5,  &;C.,  is  a  progression  by  difference, 
the  common  difference  being  1 ;  also,  3,  5,  7,  9,  11,  &c., 
the  common  difference  being  2. 

A  progression  is  called  increasing,  when  the  terms 
increase  from  left  to  right;  and  it  is  called  decreasing, 
when  the  terms  decrease  in  the  same  direction.  Thus, 
8,  11,  14,  17,  &c.,  is  an  increasing,  but  25,  20,  15,  10, 
&c.,  is  a  decreasing  progression. 

Art.  133.  To  exhibit  a  progression  by  difference  in 
its  most  general  form,  let  a  be  the  first  term,  and  d  the 
common  difference. 

Then,  if  the  progression  is  increasing, 

iBt  2d  3d  4th  5th 

a,  (a  +  ^)»  («  +  2^),  («  +  3rf),  (a  +  4rf),  &c., 
will  be  the  successive  terms  at  the  commencement  of  the 
series. 

But  if  the  progression  is  decreasing, 

Ist  2d  3d  4th  5th 

a,  (a  —  d),  {a  —  2cZ),  {a  —  ^d),  {a  —  46?,)  &/C., 
will  be  the  initial  terms. 

If  we  examine  either  of  these  series,  we  shall  perceive 
that  the  coefficient  of  d  in  the  second  term  is  1,  in  the 
third  term  it  is  2,  in  the  fourth,  3,  in  the  fiflh,  4,  &>c. ; 


<§>4*2.]  PROGRESSION    BY    DIFFERENCE.  229 

that  is,  the  coefficient  of  d  in  any  term  is  always  less  by 
1  than  the  number  which  marks  the  place  of  the  term. 
In  other  words,  to  find  any  term,  we  multiply  the  common 
difference  by  a  number  less  by  1  than  that  which  marks 
the  place  of  the  term,  and  add  the  product  to  the  first 
term  when  the  progression  is  increasing,  but  subtract  the 
product  from  the  first  term  when  the  progression  is  de- 
creasing. 

Hence  if,  in  addition  to  our  previous  notation,  we  de- 
note the  number  of  terms  by  w,  and  the  last  term  by  /,  we 
have  the  formula 

/=  «  -f-  (w  —  l)d/ui  an  increasing  progression ;  and 
l^zia —  {n  —  1)  J,  in  a  decreasing  progression. 

If  the  double  sign  zh  be  used,  the  general  formula  for 
the  last  term  is 

/=«rh(w  —  \)d.     Hence, 

To  find  the  last  term^  multiply  the  common  difference  hy 
the  number  of  terms  minus  one,  and  add  the  product  to  the 
first  term  if  the  progression  is  increasing ,  hut  subtract 
the  product  from  the  first  term  if  the  progression  is  di* 
minishing. 

1.  Required  the  12th  term  of  the  progression,  7,  10, 
13,  16,  &c. 

In  this  example,  a  z=  7,  d=zS,  and  n=^12;  and  by  sub- 
stituting these  numbers  in  the  formula,  l=ia-\-{n  —  1)  d, 
we  have 

/=:7'+(12  — l)3z=:7+11.3=:7  +  33=:40. 

Therefore,  the  12th  or  last  term  is  40. 

2.  Required  the  9th  term  of  60,  55,  50,  &c. 

In  this  example,  a=z60,  dzrz5,  and  n=:9,  and  the 
progression  is  decreasing.  Hence,  l=^a  —  {n —  1)  d  be- 
comes, by  substitution, 

i[=60  — (9—1)5  =  60  — 8.5=160  — 40  =  20. 


930  PROGRESSION    BY    DIFFERENCE.  [^42. 

Art.  lSt6.  Let  us  now  proceed  to  find  a  formula  for 
the  sum  of  any  number  of  terms.  For  this  purpose,  let 
S  represent  the  sum  of  n  terms  of  the  progression,  «, 
a  -j-  6?,  a-\-2  d,  &lc.     Then, 

1st  2d  3d  4th  nth 

^=:«+(a  +  J)  +  (a  +  26?)  +  (a  +  3^)  + +  /.(!). 

If  we  write  the  progression  in  the  reverse  order,  begin- 
ning with  the  last  term,  it  is  plain  that  the  successive 
terms  of  the  same  progression  will  be  Z,  / — d,  I — 2d, 
&/C.     Hence, 

nth  (n  — l)st  (n— 2)d  {n  — 3)d  1st 

S=zl-^{I—d)-\-{l—2d)-^{l—Sd)  + +  «.  (2). 

Remark.  It  is  manifest  that  the  terms  cannot  all  be 
written,  unless  some  determinate  value  is  given  to  n.  We 
therefore  use  points  to  supply  the  place  of  the  indefinite 
number  of  terms. 

By  adding  equations  (I)   and  (2),  and  observing  that 
d,  2d,  3c?,  &/C.  in   (1),   are   cancelled   by  — d,  — 2d, 
—  Sd,  &/C.  in  (2),  we  have 
2  S=  («+0  +  («+Z)+(a+/)+(«+Z)  + +  («+Z). 

But  since,  in  this  last  equation,  the  quantities  included 
between  the  several  parentheses  are  the  same,  and  since 
this  same  quantity  a  -{-  I  is  repeated  as  many  times  as 
there  are  terms  in  the  progression,  that  is,  n  times,  the 
second  member  is  the  same  as  n{a-\-'l).  Hence. 
2S=n{a  +  l),  .-. 

S  = .     I  his  IS  the  same  as 

2 

S=  —  {a-{-I),orn, .     Hence, 

To  Jind  the  sum  of  any  number  of  terms  in  progression 
hy  difference,  multiply  the  sum  of  the  first  and  last  terms 
by  half  the  number  of  terms,  or  multiply  half  the  sum  of 
the  first  and  last  terms  by  the  number  of  terms. 


§43.]  PROGRESSION    BY    DIFFERENCE.  231 

Required  the  sum  of  8  terms  of  the  series  6,  10, 14,  &c. 

In  this  example,  a=z  6,  d=z  4t,  and  w  =  8.     We  are 
first  to  find  the  value  of  /,  which  by  the  preceding  Article 
is  Z  =:  6  +  (8  —  1 )  4  =  34.     Then,  substituting  the  values 
of  a,  /,  and  n  in  the  formula  for  S,  we  have 
>Sf=  1(6  +  34)  =  4.  40  =160. 


SECTION   XLIII. 

EXAMPLES    IN    PROGRESSION    BY    DIFFERENCE. 

Art.  137.  1.  Required  the  12th  term  of  the  series 
10,  16,  22,  &c. 

2.  Required  the  20th  term  .  of  the  series  100,  98, 
96,  &c. 

3.  What  is  the  sum  of  100  terms  of  1,  2,  3,  4,  &c.  ? 

4.  Find  the  8th  term  and  the  sum  of  the  first  8  terms 
of  7,  10,  13,  &c. 

5.  Required  the  sum  of  10  terms  of  the  series,  in 
which  the  first  term  is  2,  and  the  common  difference  ^. 

6.  Required  the  25th  term,  and  the  sum  of  the  first  25 
terms  of  the  series  60,  59f ,  59J,  &lc, 

7.  A  man  buys  10  sheep,  giving  2  s.  for  the  first,  4  s. 
for  the  second,  6  s.  for  the  third,  and  so  on.  How  much 
do  they  all  cost  him  ? 

8.  Twenty  stones  and  a  basket  are  in  the  same  straight 
line,  and  5  yards  asunder ;  how  far  would  a  boy  travel,  if, 
starting  from  the  basket,  he  were  to  pick  up  the  stones, 
and  carry  them  one  by  one  to  the  basket  ? 

9.  Separate  39  into  three  parts  which  shall  be  in  arith- 
metical progression,  the  common  difference  being  7. 

Let  X  =z  the  least  part,  or  first  term  of  the  progression 


232  PROGRESSION    BY    QUOTIENT.  [<§>  44. 

10.  Find  three  numbers  in   arithmetical  progression 
such  that  their   sum   shall   be   30,   and    their  continued 
product  750. 

Let  y  =  the  common  difference,  and  x  -=.  the  middle 
term.  Then  x  —  y,  x,  and  x-\-y  will  represent  the 
numbers. 

11.  Two  men,  189  miles  asunder,  set  out  at  the  same 
time  to  travel  towards  each  other  till  they  meet.  One  of 
them  goes  10  miles  each  day ;  the  other  goes  3  miles  the 
first  day,  5  the  second,  7  the  third,  and  so  on.  In  how 
many  days  will  they  meet  ? 

Let  X  =:  the  number  of  days ;  then  x  will  represent 
the  number  of  terms,  and  will  correspond  to  n  in  the 
formula. 

12.  Two  travellers,  135  miles  asunder,  set  out  at  the 
same  time  to  travel  towards  each  other.  One  travels 
5  miles  the  first  day,  8  the  second,  11  the  third,  and  so 
on ;  the  other  travels  20  miles  the  first  day,  18  the  second, 
16  the  third,  and  so  on.  In  how  many  days  will  they 
meet  ? 


SECTION   XLIV. 


PROGRESSION    BY    QUOTIENT. 


Art.  138.  A  progression  by  quotient,  or  geometrical 
progression,  is  a  series  of  quantities  such,  that  if  any  one 
of  them  be  divided  by  that  which  immediately  precedes 
it,  the  quotient  will  be  the  same,  in  whatever  part  of  the 
series  the  two  quantities  are  taken.  The  successive  quan- 
tities are  called  terms  of  the  progression. 

The  quotient  arising  from  the  division  of  any  term  by 


<§>44.]  PROGRESSION    BY    QUOTIENT.  233 

that  which  immediately  precedes  it,  is  called  the  common 
ratio. 

.  For  example,  3,  6,  12,  24,  &/C.  is  a  progression  by 
quotieht,  the  common  ratio  being  2;  also,  100,  20,  4, 
t>  ih>i  ^^'  is  ^  similar  progression,  the  common  ratio 
being  |. 

A  progression  by  quotient  is  called  increasing  or  dc' 
creasing,  according  as  the  terms  increase  or  diminish 
from  left  to  right.  The  former  of  the  preceding  progres- 
sions is  increasing,  the  latter  decreasing. 

Art.  130.  In  order  to  exhibit  a  progression  by  quo- 
tient in  its  most  general  form,  let  a,  h,  c,  d,  &lq,.  represent 
the  successive  terms  at  the  commencement  of  the  series, 
and  let  q  be  the  common  ratio. 

Now,  since  from  the  definition  of  a  progression  by  quo- 
tient, each  term  is  equal  to  q  times  the  preceding  term 
we  have 

b=zaq,  c=:^aq^,  d=zaq^,  e=:aq^,  &c. 

Representing  the  last  term  by  /,  and  supplying  the 
place  of  the  indefinite  number  of  intermediate  terms  by 
dots,  the  terms  of  the  progression  will  be 

Ist      2d  3d  4th  5th  6th 

a,  aq,  a  q^,  aq^,  aq"^,  aq^, ,  /. 

We  see  th^,t  the  exponent  of  q,  in  any  term  of  this 
series,  is  less  i  y  1  than  the  number  which  marks  the  place 
of  the  term.  Thiis,  the  5th  term  is  a  q"^,  the  6th  is  a  q^, 
&/C.  Hence,  if  n  represent  the  number  "of  terms,  the  7?.th 
or  last  term  will  be  aq""'^.  But  /  also  represents  the  last 
term.     Therefore, 

This  is  the  formula  for  the  last  term.     Hence, 

To  find  any  term  of  a  progression  by  quotient,  multiply 
the  first  term  hy  that  power  of  the  common  ratio,  denoted 


»*    ^  234  PROGRESSION    BY    QUOTIENT  "7        [^44 

i> 

J  ^         bi/  a  number  less  by  1  than  that  which  marks  the  j.lace  of 
the  term, 

"^  1.    What  is  the  fifth  term  of  the  progression  5,   15^ 

K^.     45,  &c.? 

\  ^     In  this  example,  a  =15,  q=^S,  and  w  =  5 ;  hence, 

'   ^  /=«g"-i  becomes /= 5. 34  =  5. 81  zn 405. 

|>    *     2.    Required   the  seventh  term    of  the   series   12288^ 

j*        ^      In  this  case,  a  =  12288,  q=^i,  and  n=:7.    Therefore| 
^       1=  12288  .  (i)«  :zi  12288  .  ^^^  =  3.     J   _^  ^y'  '    ' 

^  Art.  14:0«     We  now  wish  to  find  a  formula  for  the 

V     ,  sum  of  any  number  of  terms  of  a  progression. 

*^  Let  S  denote  the  sum  of  any  number  n  of  terms  of  the 

.       I  series  a,  aq,  aq^^  &>c.     Then, 

>  S=a-\-aq-^aq^-\-aq^-^aq'^-\' +  «^'— ^^-a^"-!. 

Multiplying  this  equation  by  q,  we  have 
qSz=iaq-\-aq^-\'aq^-\'aq'^-\'aq^-\-....-\'aq''^^-\-aq''. 

By  observation  and  a  little  reflection  we  shall  perceive,  , 
that,  if  the  indefinite  number  of  omitted  terms  were  sup- 
plied^ the  terms  in  the  second  members  of  the  two  equa- 
tions would  all  be  alike,  with  the  exception  of  a  and  aq"". 


}  s*-  Hence,  by  subtractii^g  the  first  equation  from  the  second, 

N  '  all  the  terms  in  the  second   members  will   cancel  each 

^  other  except  these  two,  and  the  subtraction  gives 

^  qS — S^^aq"^  —  a;  or  (q  —  l)S=:^aq'^  —  a.     (Art:  59.) 
^*  Hence,  dividing  by  q  —  1,  we  have 


Such  is  the  formula  for  the  sum  of  any  number  of 


^44.J  PROGRESSION    BY    Q.UOTl£NT.  235 

terms ;  but  we  may  also  obtain  another.     In  the  preceding 
Article  we  had 

l=zaq''~^;  multiplying  by  q,  we  have 
Iqzznaq'', 
Substituting  Iq  instead  of  aj"  in 

S  =  —^ ,  we  have 

We  have  then,  for  the  sum  of  a  geometrical  progression, 
the  two  following  formulae,  viz., 
,^^ajr^^  and 
9  —  1 

9-1 

Hence, 

To  find  the  sum  of  a  progression  by  quotient ^  raise  the 
common  ratio  to  the  power  denoted  by  the  number  of  terms, 
subtract  1  from  this  power,  multiply  the  remainder  by  the 
first  term,  and  divide  the  product  by  the  ratio  minus  1 ; 
or,  multiply  the  last  term  by  the  ratio,  subtract  the  first 
term  from  the  product,  and  divide  the  remainder  by  the 
ratio  minus  1. 

1.    Required  the  sum  of  six  terms  of  the  series,  4,  8, 
16,  &c. 
In  this  example,  «  =  4,  g'  =  2,  and  w  =  6.     Hence, 

8=i becomes,  by  substitution, 

^^ii^Llll)^  1^=2)^4.  63  =  252 
2—1  1 

Or  we  may  find  the  last  term,  and  then  use  the  formula 
^_ij-a^    Bythe  last  Article, /=  4. 25  =  4.32=  28 


V 


yO-  \^     SP®/lU  PROGRESSION    BY    QUOTIENT.  [<§>  44 


Then, 


2  —  1  1 

2.    Required  the  sum  of  seven  terras  of  the  series  2,  1, 

In  this  case,  a  =  2,  g'  =  s"?  ^^^  /i  =  7.  Hence,  using 
the  first  formula  for  S,  we  have 

Art.  14:1.  Whenever,  in  the  formula  for  S,  the  ratio 
g'  is  a  proper  fraction,  that  is,  a  fraction  less  than  I,  q  —  1 
will  be  negative.  Also,  g'" — 1  will  be  negative,  because 
any  power  of  a  proper  fraction,  the  index  of  the  power 
•being  greater  than  1,  is  always  less  than  the  fraction  itself 
Thus,  {^y  :=.  -^  is  less  than  ^. 

Changing  the  signs  of  numerator  and  denominator  in 
the  formula  for  S,  which  does  not  alter  the  value  of  the 
fraction,  we  have 

Now,  since  the  powers  of  a  proper  fraction  constantly 
diminish  in  value,  as  the  exponent  of  the  power  is  in- 
creased, it  follows  that  if  w,  the  number  of  terms,  is 
infinitely  great,  g"  must  be  infinitely  small,  and  may  be 

considered    zero.      In   this    case,    S  =z —   becomes 

^       a  —  a .  0  a 

Since  q  is  supposed  to  be  a  fraction,  let  it  be  rep- 
resented by  — ,  so  that  g=z—.     Substituting  —  instead 


f^-\  ^►>^--  3?--^7  7  - 


^45.]  PROGRESSION    BY    QUOTIENT.  237 

of  q  in  the  last  formula,  we  have 

8=1  — —.     Multiplying  numerator  and  denominator  by  n, 
1 \ 

n 

n  — «i 

This  is  the  formula  for  the  sum  of  a  decreasing  pro- 
gression by  quotient,  continued  to  infinity. 

Hence,  * 

To  find  the  sum  of  an  infinite  decreasing  series  in 
progression  by  quotient,  multiply  the  first  term  by  the  de^ 
nominator,  and  divide  the  product  by  the  difference  between 
the  denominator  and  numerator  of  the  ratio, 

1.  Required  the  sum  of  the  series  7,  ^,  ff ,  &c.,  con- 
tinued to  infinity. 

In  this  example,  a  =  7,  qz=z  —  z=z—.     Hence, 

8=l^  =  f-^  =  V7X, 
5  —  3        2  ^ 


SECTION  XLV. 

EXAMPLES    IN    PROGRESSION    BY    QUOTIENT, 

Art.  143.     1.    What  is  the  sum  of  10  terms  of  the 

series,  1,  2,  4,  8,  &c.  ?  i  Q  ^  S. 

2.  Required  the  6th  term  and  the  sum  of  the  ^rst  9       ^     \ 
terms  of  the  series,  5,  20,  80,  &c.         ^  ^^i"^'    5^     '  /d*^  V^ 

3.  Required  the  sum  of  10  terms  of  the  progressioib^ 
8,  4,  2,  1,  ^,  i,  &c.  /  ^        ' 

4.  Required  the  sum  of  the  preceding  series,  continued 
to  infinity. 


338 


PROGRESSION    BY    Q,UOTIENT. 


[^45 


5.  What  is  the  sum  of  the  series,  2,  f ,  ^f ,  &,c.,  con 
tinued  to  infinity  ? 

6.  What  three  numbers  form  a  geometrical  progression, 
in  which  the  mean  term  is  8,  and  the  sum  of  the  extremes 
34? 

o 

Let  X  =1  the  ratio.     Then  — ,  8,  and  8  x  will  represent 

X 

the  terms. 

7.  Three  numbers  in  progression  by  quotient  are  such, 
that  the  sum  of  the  first  two  is  90,  and  the  sum  of  the 
'ast  two  180.     Required  these  numbers. 

Let  X  =  the  least  number,  and  y  i:=  the  ratio.     Then, 
X,  X2/,  and  xt/^  will  represent  the  numbers.     Hence, 

ix    -\-xy  =    90; 

Uy-f-^y^==180. 
Remark,     One  of  these  equations  can  be  divided  by 
the  other. 

8.  Four  numbers  are  in  geometrical  progression.  The 
sum  of  the  first  three  is  62,  and  the  sum  of  the  last  three 
is  310.     What  are  these  numbers? 

9.  Separate  105  into  three  parts  which  shall  form  a 
geometrical  progression,  such  that  Slie^  third  term  shall 
exceed  the  first  by  75.  \  /:  \" 

10.  The  sum  of  three  numbers  in\  progression  by 
quotient  is  91,  and  the  ii^an  term  is  to  the  simrjif  the 
extremes  as  3  ta  ll).     Required  these  numbei^fe"^"""^ 


,.j> 


s>-P^ 


y 


/^vi 


JO'  / 


^/;^' 


US' 


:  % 


/  D  ^ 


^('^ 


/ 


j,]^  -' 


•^■■<\X%'^'' 


,./ 


^  .v'^  ..«r 


.puJ^la^f^'^  to:^ 


o^-j-  dd 


Q^ 


-A^^'l 


/ 


-f&^'y^ 


'^ 


^ 


X    t 


3  ,-! 


\< 


y-  ^  X 


^>^3^ 


/ 


6u 
r, 


-f4  o.f^T-  ^ 


'^  ^}y    a^ 


a-i-^eJ^f^l^  ^ 


3. 


a 


^  \ 


^ 


K)! 


1u       cvj 


K 
^ 


^/j 


■6-f  -  cfioA-    -  — 


9^ — Z 

X  (5 


/4 


t 


-^ 


tx 


:j 


sk 


V 


-(- 


=,-l^ 


4 
I 


n- 


> 


"^ 


i- 


% 


l;i 


T 


1  -^"^ 


(V 


Cs) 


\ 


ci^         «^        V) 


A» 


,§         ^        ^^       ^       V  * 


Oo  "^ 


w 


t3 


.:%x-) 


n 


?i 


^-^r 


P306125 


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